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AN  ALGEBRAIC  ARITHMETIC 


» 


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AN  ALGEBRAIC  ARITHMETIC 


AN  EXPOSITION   OF  THE  THEORY 
AND  PRACTICE   OF 

ADVANCED   ARITHMETIC 

BASED  ON  THE  ALGEBRAIC  EQUATION 


BY 

S.  E.  COLEMAN,  B.S. 

WILLIAM  WHITING  FELLOW  AT  HARVARD    UNIVERSITY 

FORMERLY  INSTRUCTOR  IN  MATHEMATICS  IN  THE  OAKLAND 

HIGH   SCHOOL,   OAKLAND,  CALIFORNIA 


Wetjj  gorit 
THE    MACMILLAN   COMPANY 

LONDON :  MACMILLAN  A  CO.,  Ltd. 
1898 

All  rights  reserved 
/      *■'     OF  THE 

UNIVERSITY 

OF 

£^LlFORfi:^ 


Copyright,  1897, 
By  the  MACMILLAN  COMPANY. 


Vodnooti  ^xt»» 

J.  S.  Cuihinj?  A  Co.  -  Rerwick  k  Smith 
Norwood  MasB.  U.S.A. 


PREFACE 

This  arithmetic  is  not  ojffered  to  the  public  as  a  refine- 
ment or  super-refinement  of  the  methods  of  existing  text- 
books on  the  subject.     It  is  a  new  departure. 

For  a  number  of  years  arithmetics  have  been  under- 
going a  progressive  change.  Voluminous  works,  in  which 
the  isolated  treatment  of  related  topics  and  the  multi- 
plicity of  detail  relating  to  business  arithmetic  completely 
obscured  the  unity  of  the  science,  have  by  degrees  given 
place  to  more  compendious  works.  The  change  has, 
however,  been  little  more  than  a  process  of  successive 
elimination.  To  the  former  plethora  has  succeeded  an 
ever-increasing  leanness,  until  at  last  the  skeleton  of  the 
subject  stands  revealed  indeed,  for  it  alone  remains.  The 
arithmetic  of  to-day  is  merely  a  compilation  of  examples, 
classified  and  miscellaneous,  with  illustrative  solutions 
accompanied  by  brief  explanatory  notes  and  a  few  defini- 
tions. The  task  of  infusing  a  living,  rational  principle 
into  these  dry  bones  is  left  entirely  to  the  teacher. 

A  second  change  in  the  mathematics  of  the  grammar 
school,  contemporaneous  with  that  above  mentioned,  has 
been  the  introduction  of  elementary  geometry  and,  more 
recently,  of  elementary  algebra.  The  situation  is  best 
described  by  saying  that  the  latter  subjects  have  partially 


173093 


VI  PREFACE 

superseded  arithmetic,  since  the  whole  time  devoted  to 
mathematics  has  remained  substantially  the  same. 

The  reason  for  these  changes  is  not  far  to  seek.  The 
mathematics  are  chiefly  valuable  as  a  factor  in  education 
in  that  they  afford  a  means  of  developing  the  reasoning 
powers  of  the  child;  and  as  arithmetic,  in  spite  of  the 
numerous  attempts  to  improve  the  text-books  and  the 
methods  of  teaching,  persisted  in  remaining  little  more 
than  a  collection  of  rule-of-thumb  methods  for  turning 
out  "answers,"  the  progressive  teacher  naturally  turned 
to  other  branches  of  mathematics  which  embodied  a  logi- 
cally coherent  science  not  yet  perverted  by  "practical 
applications." 

Much  was  gained  by  so  doing.  Geometry,  which  in 
its  complete  and  rigidly  demonstrative  form  is  a  fairly 
difficult  subject  for  the  high  school,  was  found  to  contain 
a  large  number  of  facts  that  could  be  established  by 
simple  yet  fairly  conclusive  reasoning.  And  not  only  is 
the  method  by  which  these  facts  are  acquired  in  their 
logical  relation  of  the  highest  value  in  developing  the 
reasoning  powers  of  the  child,  but  the  facts  themselves 
possess  a  value  far  higher  than  that  of  mere  utility.  For 
example,  it  may  possibly  prove  of  use  to  some  member  of 
a  large  grammar-school  class  to  know  the  empirical  rule 
by  which  the  contents  of  a  cask  or  the  number  of  feet  of 
lumber  in  a  round  log  are  determined ;  but  it  is  important 
that  all  should  know,  and  should  be  able  to  give  some 
simple  explanation  of  the  fact,  that  similar  surfaces  are 
to  each  other  as  the  squares  and  similar  solids  as  the 
cubes  of  their  like  dimensions.  These  are  universal 
truths,  depending  on  the  nature  of  space,  by  which  all 


PREFACE  Vll 

physical  existence  is  conditioned;  and  a  knowledge  of 
them  is  therefore  an  essential  part  of  a  complete  educa- 
tion; but  this,  from  its  very  nature,  can  be  true  of  no 
empirical  rule. 

The  introduction  of  elementary  algebra  into  the  schools 
is  of  more  doubtful  value  for  two  principal  reasons. 
The  conceptions  of  geometry  can  be  represented  by  fig- 
ures and  objects,  and  are  therefore  readily  grasped  by 
the  child,  but  those  of  algebra  can  be  so  represented  only 
to  a  very  limited  degree;  and,  for  the  most  part,  are 
abstractions  the  nature  of  which  the  child  comprehends 
with  difficulty.  Moreover,  the  time  that  can  be  allotted 
to  the  subject  in  the  grammar  school  is  barely  sufficient 
to  carry  the  pupil  over  the  essentially  uninteresting 
details  of  algebraic  manipulation  that  necessarily  precede 
any  but  the  simplest  applications  of  the  science. 

But  while  the  wisdom  of  introducing  elementary  algebra 
into  the  schools  may,  for  these  and  minor  reasons,  be 
seriously  questioned,  the  experience  of  several  years  as 
a  teacher  has  led  the  author  to  the  conclusion  that  the 
application  of  certain  algebraic  conceptions  to  arithmetic 
would  contribute  largely  toward  the  rational  presentation 
of  the  subject,  thus  increasing  its  disciplinary  value,  and 
at  the  same  time  preparing  the  way  for  a  natural  transi- 
tion to  the  algebra  of  the  high  school. 

These  conceptions  are  the  use  of  letters  as  the  general 
representatives  of  (positive)  numbers  and  of  the  equation 
to  express  their  relations. 

Both  ideas  are  introduced  into  the  first  chapter,  and 
developed,  so  far  as  the  purpose  demands,  in  the  second. 
The  comparatively  large  amount  of  space  given  to  prob- 


Vlll  PREFACE 

lems  is  due  to  the  fact  that  they  afford  at  once  the  most 
interesting  introduction  to  the  subject  and  the  best  means 
of  explaining  the  significance  of  the  equation  and  its 
transformation.* 

Among  the  numerous  and  important  applications  of 
these  ideas  throughout  the  book,  may  be  noted  the 
following : 

The  cases  of  percentage  are  reduced  to  three  (Art.  32), 
and  all  are  shown  to  be  contained  in  the  single  equation 
p  =  hr. 

All  the  applications  of  percentage  not  involving  time 
are  shown  to  be  merely  special  applications  of  the  three 
percentage  formulas.  (For  example,  see  table  under 
Profit  and  Loss,  Art.  33.) 

All  the  formulas  of  simple  interest  are  derived  from 
the  two:  i=prt  and  a=p-\-i-j  these  being  obtained 
directly  from  the  definitions. 

The  interest  formulas  are  shown  to  be  special  develop- 
ments of  the  earlier  percentage  formulas  (Art  48). 

The  principles  of  proportion  are  rigidly  demonstrated, 
affording  a  simple  and  elegant  illustration  of  the  method 
of  algebraic  proof,  freely  used  in  the  chapter  on  men- 
suration. 

*  The  attention  of  teachers  is  called  to  the  fact  that  an  equation 
is  not  a  quantity,  but  an  expression  of  relation,  and  therefore  can- 
not be  operated  upon,  in  the  usual  sense  of  the  word.  Operations 
are  not  performed  upon  an  equation,  but  upon  its  members.  The 
abbreviated  and,  to  beginners,  highly  misleading  forms  of  state- 
ment, "  Multiply  the  equation  by  3,"  "Subtract  10  from  the  equa- 
tion," and  the  lik«,  should  be  studiously  avoided.  For  a  similar 
reason,  the  word  "transpose"  should  not  be  used.  A  term  is 
transposed  by  addition  or  subtraction,  and  the  specific  operation 
should  always  be  named. 


PREFACE  IX 

The  algebraic  and  geometrical  explanation  of  evolution 
are  combined  in  one,  the  algebraic  symbols  being  the 
natural  method  of  expressing  the  geometrical  relations. 
Thus  each  set  of  ideas  confirms  the  other. 

The  chapter  on  mensuration  embodies  the  principles 
enunciated  in  the  early  part  of  this  preface  as  fully  as 
the  limited  time  generally  allotted  to  this  subject  permits. 
In  schools  where  a  more  extended  course  in  elementary 
geometry  is  given,  it  will  afford  a  convenient  opportunity 
for  a  review  of  the  most  important  results  of  the  course. 
The  author  takes  pleasure  in  acknowledging  his  indebted- 
ness to  Hill's  Lessons  in  Geometry  for  many  valuable 
suggestions  in  the  preparation  of  this  chapter,  and  recom- 
mends the  book  as  being  admirably  adapted  to  the  needs 
of  grammar  schools. 

Among  the  features  of  the  book  not  resulting  from  the 
algebraic  method  of  treatment,  the  author  would  call 
attention  to  the  use  of  the  article  to  mark  the  logical 
divisions  of  the  subject;  to  the  rational  explanation  of 
the  application  of  simple  and  compound  proportion  to  the 
solution  of  problems;"*  and  to  the  treatment  of  partial 
payments.  The  latter  subject  is  placed  after  compound 
interest,  where  the  effect  of  the  two  usual  methods  of 
applying  partial  payments  can  be  intelligibly  discussed, 
and  the  manner  in  which  interest  is  compounded  by  the 
United  States  Rule  is  fully  explained. 

It  is  believed  that  the  examples,  which  for  the  most 
part  have  been  compiled  from  various  sources,  present  a 

*  The  author  has  never  seen  anything  on  this  point  in  any  text- 
book, except  variations  of  the  rule  of  thumb  :  More  requires  more, 
and  less  requires  less. 


X  PREFACE 

fairly  extensive  and  varied  application  of  the  subject 
matter. 

The  author  will  receive  with  pleasure  any  suggestions 
or  criticisms  that  will  be  of  assistance  in  the  improvement 
of  later  editions  of  this  work. 

S.  E.  COLEMAN. 

Cambbidgb,  Mass.,  Aug.  11, 1897. 


CONTENTS 

CHAPTER  I 
INTRODUCTION 

PAOB 

The  Usb  of  Letters  to  denote  Numbebs  ....  1 

The  Equation        . 4 

Definitions 6 

Axioms 7 

The  Solution  of  Equations 7 

The  Solution  of  Pboblems 9 

Directions  for  the  Solution  of  Problems       ...  12 

CHAPTER  II 

ADDITION.    SUBTRACTION.     MULTIPLICATION.    DIVISION 

Inverse  Processes 15 

Addition 16 

Subtraction 17 

Multiplication 20 

Division 27 

CHAPTER  III 
PERCENTAGE  AND  ITS  APPLICATIONS 

Percentage 31 

Percentage  Formulas 37 

xi 


XU  CONTENTS 

PAQB 

Profit  and  Loss 38 

Commission  and  Brokerage 43 

Commercial  Discount 46 

Insurance 48 

Taxes 50 

Duties 62 

CHAPTER  IV 

APPLICATIONS  OF  PERCENTAGE  INVOLVING   TIME 

Simple  Interest 57 

Accurate  Interest 62 

Problems  in  Interest 63 

Interest  Formulas 68 

Present  Worth  and  True  Discount 70 

Bank  Discount 72 

Annual  Interest 75 

Compound  Interest 76 

Partial  Payments         .        . 78 

CHAPTER  V 
PROPORTION.     PARTNERSHIP.     AVERAGE  OF  PAYMENTS 

Ratio 84 

Proportion 85 

Problems  in  Simple  Proportion 89 

Compound  Proportion 92 

Problems  in  Compound  Proportion 94 

Partnership .        .97 

AyBRAOB  of  Payments .        .  99 


CONTENTS  XUl 

CHAPTER   VI 
INVOLUTION  AND  EVOLUTION 

PAOE 

Involution 101 

Evolution 102 

The  Square  of  the  Sum  of  Two  Numbers        .        .         .  103 

Square  Root 105 

The  Cube  of  the  Sum  of  Two  Numbers    .         .        .        .110 

Cube  Root 112 

CHAPTER  VII 
MENSURATION 

Lines 117 

Angles 118 

Plane  Figures 118 

Area  of  Parallelograms 120 

Area  of  Triangles 121 

Area  of  Polygons .  124 

The  Circle 126 

Similar  Plane  Figures 131 

Solids 132 

Prisms  and  Cylinders 132 

Pyramids  and  Cones 135 

The  Sphere 139 

Similar  Solids 141 


ALGEBRAIC   ARITHMETIC 

CHAPTER  I 

INTRODUCTION 

1.  The  Use  of  Letters  to  denote  Numbers.  It  is  often 
necessary  to  speak  of  something  whicli  is  true  not  only 
of  one  number  or  set  of  numbers,  but  of  all  numbers  or 
of  all  similar  sets  of  numbers.  For  example,  the  sum  of 
9  and  6  is  15,  and  their  difference  is  3.  If  we  add  this 
sum  and  difference,  we  get  18,  which  is  twice  the  greater 
of  the  two  numbers.  If  we  subtract  the  difference  from 
the  sum,  we  get  12,  which  is  twice  the  smaller  of  the  two 
numbers.  The  same  relation  is  true  of  any  two  numbers, 
and  may  be  expressed  in  the  general  statement : 

If  the  sum  and  difference  of  two  numbers  he  added,  the 
result  is  twice  the  greater  of  the  numbers;  if  the  difference 
be  taken  from  the  sum,  the  result  is  twice  the  smaller  of  the 
numbers. 

This  may  be  expressed  more  briefly  by  the  aid  of 
signs;  thus: 

sum  of  two  numbers  +  difference  of  the  numbers 

=  twice  the  larger  number. 

sum  of  two  numbers  —  difference  of  the  numbers 

=  twice  the  smaller  number. 

B  1 


2  ALGEBRAIC   ARITHMETIC 

The  statement  may  be  still  further  shortened  by  using 
letters  to  represent  the  numbers.  Thus,  let  g  stand  for 
the  greater  number,  I  for  the  less,  s  for  their  sum,  and 
d  for  their  difference.     We  then  have 

s-^d  =  2g, 
and  s  —  d  =  2l; 

in  which,  it  must  be  remembered,  a  and  b  denote  any 
two  numbers,  provided  only  that  a  is  greater  than  b. 
From  the  above  equations  it  is  clear  that 

9='-ii  (1) 

and  '  =  '-^-  (2) 

These  equations  express  the  fact  that  the  larger  of  two 
numbers  is  equal  to  one-half  the  sum  found  by  adding  the 
sum  and  the  difference  of  the  numbers,  and  the  smaller 
number  is  equal  to  one-half  the  remainder  found  by  sub- 
tracting the  difference  of  the  numbers  from  their  sum. 
These  equations  are,  in  fact,  a  very  convenient  state- 
ment of  the  rule  for  finding  any  two  numbers  when 
their  sum  and  difference  are  given. 

Note.  Rules  stated  in  the  form  of  equations  are  called  for- 
mulas. 

The  following  chapter  will  afford  numerous  illustra- 
tions of  the  advantage  of  using  letters  to  denote  numbers 
when  we  are  studying  their  general  properties. 

2.  Operations  to  be  performed  with  numbers  denoted 
by  letters  are  indicated  by  the  usual  signs  of  arithmetic, 
in  the  same  manner  as  when  the  numbers  are  expressed 


INTRODUCTION  8 

with  figures ;  with  the  exception  that  the  product  of  two 
numbers  is  indicated  by  writing  the  letters  together,  the 
sign  of  multiplication  being  omitted. 

Thus  the  sum  of  any  two  numbers  a  and  b  is  indicated 
by  a  -f  6,  their  difference  by  a  —  6,  their  product  by  a6, 

and  their  quotient  by  a  ^  6,  or  by  -• 

b 

In  each  case  the  numerical  value  of  the  result  can  be 
found  only  when  the  values  of  a  and  b  are  given. 

Note  1.  The  word  numerical  relates  to  particular  numbers, 
that  is,  to  numbers  expressed  by  figures  ;  the  word  literal,  to  num- 
bers expressed  by  letters.  Thus  12  is  a  numerical  quantity  ;  a,  b, 
c,  etc. ,  are  literal  quantities. 

Note  2.  The  sign  of  multiplication  cannot  be  omitted  between 
numerical  factors,  but  is  omitted  between  a  numerical  and  a  literal 
factor.  Thus  6  x  a  x  6  is  written  Bab.  (Omit  "times"  in  read- 
ing.) 

EXAMPLES  1 

If  a  =  4,  6  =  1,  c  =  3,  and  d  =  2,  find  the  numerical 
values  of 

1.  a 4- 6.  6.   ac  —  2bd. 

2.  a  — &4-C.  7.   4:Cd-\-ab. 

3.  5a-2d.  8.   Sa-5b-\-2acd. 

4.  12b-2a.  9.   abc -^ abd  +  bed. 

5.  a  +  86  — 6d  10.   2a-r-d  +  4c. 

If  a  =  6,  6  =  5,  c  =  2,  and  d  =  0,  find  the  values  of 

11.  2ab-{-b  —  cd.  ^^    ^4.^. 

r  '   c     b 

12.  a  -T-  c  X  6  —  ac.  ^  r  ,  o  ^ 

-_    2 ac  —  6  -f  3 d 

13.  6  a6 -J- 5  c.  •       7aH-a6c     * 


4  ALGEBRAIC   ARITHMETIC 

3.  The  Equation.  The  statement  that  two  numbers  or 
two  sets  of  numbers  are  equal  is  called  an  equation. 

Equations  are  used  in  Art.  1,  and  the  values  of  the 
letters  in  the  examples  of  Art.  2  are  given  by  means 
of  equations. 

The  part  of  an  equation  on  the  left  of  the  sign  of 
equality  is  called  the  left  side,  left  member,  or  first  mem- 
ber of  the  equation;  that  on  the  right,  the  right  side, 
right  member,  or  second  member. 

Many  problems  can  be  most  easily  solved  by  the  use 
of  letters  to  denote  the  numbers  to  be  found,  and  equa- 
tions to  express  the  relations  that  exist  between  these 
numbers  and  the  given  numbers  of  the  problem.  How 
this  is  done  will  be  shown  by  the  following  examples : 

Ex.  1.  If  5  be  added  to  3  times  a  certain  number, 
the  result  is  29.     Eind  the  number. 

The  problem  may  be  stated  more  briefly  thus  : 

3  times  a  certain  number  4-  5  =  29 ; 

or,  if  we  let  a  stand  for  the  number,  it  may  be  stated 
still  more  briefly  by  the  equation 

3a  +  5  =  29.  (1) 

Now  if  from  this  equation  we  can  find  the  value  of  a, 
that  is,  the  number  that  a  represents,  this  value  will  be 
the  answer  to  the  problem.     Let  us  try  to  do  this. 

Subtracting  5  from  both  members  of  the  equation, 

we  get 

3a  =  24.  (2) 

Dividing  the  sides  of  this  equation  by  3,  we  have 

a  =  8.  (3) 


THE  EQUATION  O 

To  prove  the  result,  replace  a  in  equation  (1)  by  its 
value.    This  gives 

3x8+5  =  29, 
29  =  29. 

The  equation  is  said  to  be  satisfied  by  a  =  8 ;  which 
means  that  when  8  is  substituted  for  a,  the  equation  is 
true.  It  is  evident  that  it  would  not  be  satisfied  by  any 
other  value  of  a. 

Ex.  2.  The  sum  of  two  numbers  is  38,  and  their 
difference  is  8.     What  are  the  numbers  ? 

The  answer  can  be  written  down  at  once  by  substi- 
tuting s  =  38,  and  (Z  =  8  in  formulas  (1)  and  (2),  Art.  1. 
The  pupil  should  carefully  compare  this  method  of  solu- 
tion with  the  following : 

Let  us  denote  the  greater  of  the  numbers  by  x ;  then, 
since  the  smaller  number  is  8  less  than  the  larger,  it  will 
be  denoted  by  a;  —  8. 

The  problem  states  that 

the  larger  number  +  the  smaller  number  =  38. 
Hence  x-\-{x-^)  =  38,  (1) 

or  a;  -h  a;  -  8  =  38, 

or  2a;-8  =  38. 

Add  8  to  both  sides  of  the  equation ;  then 

2a;  =  46. 
Divide  both  sides  by  2 ;  then 

a;  =  23  =  larger  number. 
Hence       a;  —  8  =  16  =  smaller  number. 


6  ALGEBBAIC   ARITHMETIC 

Proof.  Substituting  these  values  in  equation  (1),  we 
have 

23  +  15  =  38, 

38  =  38. 

Note.  It  is  necessary  to  notice  the  punctuation  after  the  equa- 
tions in  the  solution  of  a  problem  ;  for  the  equations  always  occur 
as  parts  of  sentences,  and  the  punctuation  helps  to  make  the  mean- 
ing clear,  just  as  in  the  case  of  any  other  kind  of  sentence. 

4.  Definitions.  The  figures,  letters,  and  signs  used  in 
arithmetic  are  called  symbols. 

Any  combination  of  symbols  denoting  a  number  is 
called  an  expression.  If  it  contains  letters,  it  is  called 
an  algebraic  expression. 

The  members  of  an  equation  are  expressions. 

The  parts  of  an  expression  which  are  separated  from 
each  other  by  the  signs  of  addition  or  subtraction  are 
called  the  terms  of  the  expression. 

Thus  the  expression  2ab  ~c-\-  5  has  three  terms ; 
5(iocy  has  one  term. 

A  term  may  consist  of  two  or  more  factors.  Thus  the 
term  5  axy  contains  four  factors. 

If  the  factors  of  a  product  are  separated  into  groups 
in  any  way,  either  group  of  factors  is  called  the  coefficient 
of  the  other  group. 

Thus  in  the  term  5  axy^  5  is  the  coefficient  of  axy,  5  a 
is  the  coefficient  of  xy,  5  ay  is  the  coefficient  of  x,  etc. 
If  a  term  has  a  numerical  factor,  it  is  generally  spoken 
of  as  the  coefficient  of  the  term. 

Terms  containing  the  same  literal  factors  are  called 
like  terms. 


AXIOMS  7 

Thus  5  ahx  and  9  dbx  are  like  terms ;  3  ab  and  7  cdx 
are  unlike  terms. 

5.  Axioms.  We  have  seen  that  some  problems  can  be 
stated  in  the  form  of  equations  in  which  a  letter  stands 
for  the  answer ;  and  that  the  value  of  the  letter  which 
satisfies  the  equation  is  the  answer  to  the  problem.  In 
solving  such  equations,  frequent  use  is  made  of  the  fol- 
lowing simple  truths,  or  axioms : 

Ax.  1.  If  equal  numbers  are  added  to  equal  numbers, 
the  sums  are  equal. 

Ax.  2.  If  equal  numbers  are  subtracted  from  equal 
numbers,  the  remainders  are  equal. 

Ax.  3.  If  equal  numbers  are  multiplied  by  equal  num- 
bers, the  products  are  equal. 

Ax.  4.  If  equal  numbers  are  divided  by  equal  num- 
bers, the  quotients  are  equal. 

Thus,  if  a  =  5  and  c  =  d, 

then  a-\-  c=b-}-d  by  Ax.  1. 

a  —  c  =  b  —  d  by  Ax.  2. 

ac  =  bd  by  Ax.  3. 

and  a-i-c  =  b-i-d  by  Ax.  4. 

The  four  axioms  may  be  summed  up  in  the  statement : 
Equal  numbers  will  still  remain  equal  numbers  after  they 
have  been  increased)  diminished,  multiplied,  or  divided  by 
equal  numbers. 

6.  The  Solution  of  Equations.  Let  us  now  look  again 
at  the  solution  of  Ex.  1,  Art.  3.     The  algebraic  (or  sym- 


a  ALGEBRAIC   ARITHMETIC 

bolical)  statement  of  the  problem  is  3  a  +  5  =  29,  in 
which  a  stands  for  the  answer. 

Since  the  members  of  this  equation  are  equal  numbers, 
if  we  subtract  5  from  each  of  them,  the  remainders  will 
be  equal  by  Ax.  2.     This  gives 

3a  +  5-5  =  29-5, 
or  3a  =  24; 

that  is,  we  form  an  equation  out  of  the  equal  remainders. 
This  could  not  have  been  done  if  we  had  subtracted 
more  from  one  member  than  from  the  other,  for  in  that 
case  the  remainders  would  have  been  unequal,  and  the 
equation  would  have  been  destroyed. 

We  wish  to  obtain  a  alone  in  the  left  member  of  the 
equation.  We  can  now  do  this  by  dividing  that  side  by 
the  coefficient  of  a;  but  since  we  must  preserve  the 
equality  of  the  members,  we  divide  both  by  3,  and  obtain 
a  =  8.     In  this  operation  we  use  Ax.  4. 

Since,  in  solving  the  original  equation,  we  have  made 
use  of  only  those  operations  which  do  not  destroy  the 
equality  of  its  members,  we  know  that  the  last  equation 
is  true.     It  therefore  gives  us  the  required  value  of  a. 

Exercise.  Find  what  axioms  have  been  used  in  the 
solution  of  Ex.  2,  Art.  3. 

Ex.  1.   Find  the  value  of  x  if 

2  a;  4-  5  =  15  —  a;. 

Since  we  wish  to  obtain  x  alone  in  the  left  side,  and 
only  numerical  quantities  in  the  other,  we  must  get  rid 
of  the  X  in  the  right  side  and  the  5  in  the  left.     The  x 


THE  SOLUTION   OF   EQUATIONS  9 

will  disappear  from  the  right  side  if  we  add  x  to  it,  since 
15  —  a;  -f  a;  =  15.  Hence,  adding  x  to  both  members  to 
preserve  their  equality,  we  have 

2a;-f-5-}-a;  =  15  —  aj  +  a;  by  Ax.  1. 

or  3a;  +  5  =  15. 

Subtract  5  from  both  sides ;  then 

3a;  =  10  by  Ax.  2. 

Divide  both  sides  by  3 ;  then 

a;  =  JJ>_  =  3 J  by  Ax.  4. 

Note.  This  value  of  x  satisfies  not  only  the  given  equation, 
but  also  all  the  equations  derived  from  it ;  that  is,  x  has  the  same 
value  throughout  the  solution,  which  must  be  the  case  in  the 
solution  of  any  equation. 


EXAMPLES 

2 

Solve  the  following  equations : 

1. 

3a;  4-4: 

=  a;  + 10. 

1      7 

2. 

4a;4-4: 

=  a;  +  7.                   «• 

S^x 

3. 
4. 

5. 

5a;-5: 

a;  +  4  = 

14  =  1. 

X 

=  20-2a;.               ^ 
2(5 -a;). 

8. 

a;  +  |a; 
5^4 

=  10. 

:1. 

7.  The  Solution  of  Problems. 

Ex.  1.   What  number  is  that  whose  half  added  to  16 
gives  25  ? 

Let  X  denote  the  number. 

Then  ^  will  denote  half  the  number,  and  ^  + 16  will 
2  ^ 

denote  the  half  added  to  16. 


10  ALGEBRAIC   AKITHMETIC 

But  the  problem  states  that  this  is  25 ;  hence 
^  +  16  =  25. 

Subtract  16  from  both  sides ;  then 

1  =  9  by  Ax.  2. 

Multiply  both  sides  by  2 ;  then 

a;  =  18  by  Ax.  3. 

Proof  :  J^  +  16  =  25, 

25  =  25. 

Ex.  2.  A  man  having  $92  spent  a  part  of  it,  and  then 
had  3  times  as  much  as  he  had  spent.  How  much  did 
he  spend? 

Let  X  be  the  number  of  dollars  he  spent. 
Then  92  —  a;  will  be  the  number  of  dollars  he  had  left. 
But  the  problem  tells  us  that  this  is  3  times  as  much 
as  he  spent.     Hence  we  have  the  equation 

3a;  =  92-a;.  (1) 

Add  X  to  both  sides ;  then 

4a;  =  92  by  Ax.  1. 

Divide  both  sides  by  4 ;  then 

a;  =  23  by  Ax.  4. 

Hence  the  man  spent  %  23. 

Remarks.  In  problems  involving  concrete  numbers, 
like  the  last,  it  is  not  necessary  to  express  the  kind  of 
unit  in  the  equation.  Thus,  in  this  problem,  we  do  not 
write 

^3a;=$92~$aj; 


THE  SOLUTION   OF  PROBLEMS  11 

for,  though  the  statement  is  correct,  it  is  not  so  simple 
as  when  made  without  the  sign. 

The  members  of  the  equation  are  to  be  regarded  as 
abstract  numbers,  denoting  the  number  of  times  the  con- 
crete unit  is  contained  in  the  quantities  to  be  compared. 
Thus  the  members  of  (1)  denote  the  number  of  times  $  1 
is  contained  in  the  sum  of  money  the  man  had  left. 

The  quantities  to  be  compared  must  be  of  the  same 
kind,  and  must  be  measured  by  the  same  unit.  For 
example,  we  cannot  compare  a  sum  of  money  with  a  dis- 
tance, nor  can  we  compare  two  sums  of  money  when 
one  is  measured  in  dollars  and  the  other  in  cents  or  in 
dimes. 

Such  statements  as 

100^=  $1, 

16  oz.  =  1  lb., 

are  not  equations  at  all  in  the  sense  in  which  we  shall 
use  the  word  in  this  book.  The  first  of  these  statements 
means  that  the  two  sums  are  equal  in  value;  the  second, 
that  the  two  weights  are  equal ;  but  in  neither  of  them 
are  the  two  numbers  equal.  Equations,  as  we  shall  use 
them,  will  always  mean  that  the  two  members  are  equal 
numbers. 

Ex.  3.  A  can  do  a  piece  of  work  in  10  days,  but  A 
and  B  working  together  can  do  it  in  6  days.  In  how 
many  days  can  B  do  it  alone  ? 

Let  X  =  the  number  of  days  it  would  take  B  to  do  the 
work  alone. 

Then  -  =  the  part  of  the  work  he  can  do  in  one  day. 


12  ^  ALGEBRAIC   ARITHMETIC 

From  the  problem  we  know  that  A  can  do  ^  of  the 
work  in  one  day ;  and  A  and  B  together,  ^  of  the  work 
in  one  day. 

H+ro  (^) 

Multiply  both  sides  by  30  x,  the  L.  C.  M.  of  the  denom- 
inators; then 

20f  =  30^  +  ^  by  Ax.  3. 

6  «        10 

or  5x  =  S0-{-3x. 

Subtract  3  x  from  both  sides ;  then 

2a;=30  by  Ax.  2. 

Divide  both  sides  by  the  coefficient  of  a?;  then 

■   a;  =  15  by  Ax.  4. 

Hence  B  can  do  the  work  in  15  days. 

Proof  by  substitution :  Keplace  x  in  (1)  by  16 ;  then 

Proof  by  analysis:  Since  A  can  do  the  work  in  10 
days,  in  one  day  he  can  do  J^  of  it ;  since  B  can  do  the 
work  in  15  days,  in  one  day  he  can  do  yV  of  it.  Hence 
A  and  B  working  together  can  do  ^  4-  J^-,  or  ^,  of  it  in 
one  day,  or  the  whole  piece  of  work  in  6  days. 

8.  From  the  examples  of  Art.  3  and  Art.  7,  the  follow- 
ing directions  for  the  solution  of  similar  problems  may 
be  deduced : 

I.  Denote  the  required  number  by  some  letter  (it  is 
customary  to  use  x).    This  is  called  the  unknown  quantity. 


PROBLEMS  13 

II.  If  there  are  other  numbers  in  the  problem  that 
depend  on  the  unknown  quantity,  find  expressions  for 
them  in  terms  *  of  the  unknown  quantity. 

III.  Write  these  expressions  in  the  form  of  an  equa- 
tion which  expresses  in  symbolic  form  the  conditions 
of  the  problem. 

IV.  Clear  the  equation  of  fractions,  if  there  are  any, 
by  multiplying  both  members  by  the  /.  c.  m.  of  the 
denominators. 

V.  By  addition  or  subtraction  remove  all  terms  con- 
taining the. unknown  quantity  to  one  side  of  the  equation, 
and  all  other  terms  to  the  other  side. 

VI.  After  adding  together  the  terms  containing  the 
unknown  quantity,  divide  the  members  of  the  equation 
by  its  coefficient.     This  gives  the  answer. 

EXAMPLES  3 

1.  John  is  3  times  as  old  as  James,  and  the  sum  of 
their  ages  is  16  years.     What  is  the  age  of  each  ? 

2.  A  boy  bought  a  top  and  a  ball  for  24  cents,  paying 
5  times  as  much  for  the  ball  as  for  the  top.  What  did 
he  pay  for  each  ? 

3.  Ida's  sister  gave  her  some  money,  and  her  brother 
gave  her  twice  as  much.  After  spending  12  cents,  she 
had  18  cents  left.     How  much  was  given  her  by  each  ? 

*  A  number  is  said  to  be  expressed  in  terms  of  another  number 
when  the  expression  for  it  contains  the  letter  that  represents  the 
other  number.  Thus  in  Ex.  2,  Art.  3,  the  smaller  number,  cc  —  8, 
is  expressed  in  terms  of  the  larger  number,  x  ;  and  in  Ex.  2, 
Art.  7,  the  number  of  dollars  the  man  had  left,  3aj,  is  expressed  in 
terms  of  the  number  of  dollars  he  spent,  x. 


14  ALGEBRAIC   ARITHMETIC 

4.  The  sum  of  two  numbers  is  50,  and  their  difference 
is  18.     Find  them. 

5.  The  sum  of  three  numbers  is  126.  The  second  is 
twice  the  first,  and  the  third  is  equal  to  the  sum  of  the 
other  two.     What  are  the  numbers  ? 

6.  A  boy,  after  spending  half  his  money,  earned  14 
cents,  and  then  had  30  cents.  How  much  had  he  at 
first? 

7.  A  and  B  together  can  do  a  piece  of  work  in  8  da., 
and  A  working  alone  can  do  it  in  20  da.  In  what  time 
can  B  do  it  ? 

\/s.   Fred  has  3  times  as  many  marbles  as  Harry,  lack- 
ing 2 ;  and  both  together  have  26.    How  many  has  each  ? 

9.  The  sum  of  two  numbers  is  62,  and  the  greater  is 
3  less  than  4  times  the  smaller.     Find  the  numbers. 

10.  A  father  is  6  years  more  than  4  times  as  old  as 
his  son,  and  the  sum  of  their  ages  is  71  years.  Find 
the  age  of  each. 

11.  If  ^  of  a  certain  number  be  subtracted  from  f  of 
it,  the  remainder  will  be  8.     What  is  the  number  ? 

12.  Divide  42  into  two  parts,  such  that  one  part  shall 
be  I  of  the  other. 

13.  One  of  two  apple  trees  bore  f  as  many  apples  as 
the  other,  and  both  yielded  21  bu.  How  many  bushels 
did  each  yield  ?  ^ 

14.  A  lad  having  45  cents  bought  an  equal  number 
of  pears,  oranges,  and  bananas ;  the  pears  being  3  cents 
each,  the  oranges  4  cents,  and  the  bananas  2  cents.  How 
many  of  each  did  he  buy  ? 


CHAPTER   II 

ADDITION,  SUBTRACTION,  MULTIPLICATION,  DIVISION 

9.  We  shall  now  study  a  little  more  fully  the  way  in 
which  operations  are  performed  upon  numbers  denoted 
by  letters. 

There  are  four  fundamental  operations,  or  processes, 
by  certain  combinations  of  which  all  the  problems  of 
arithmetic  are  solved.  These  are  addition,  subtraction, 
multiplication,  and  division.  You  have  already  learned 
that  subtraction  is  the  inverse  of  addition,  and  division 
the  inverse  of  multiplication;  by  which  is  meant  that 
subtraction  itii-does  what  addition  does,  and  division  un- 
does what  multiplication  does. 

Hence,  if  to  any  number  I  add  any  other  number,  and 
afterwards  subtract  the  same  number,  I  shall  have  left 
the  first  number  unchanged,  since  the  two  operations 
exactly  cancel  each  other. 

Thus,  if  a  and  h  are  any  two  numbers, 

a  +  b-b  =  a.  (1) 

Again,  if  I  multiply  any  number  by  any  other  number, 
then  divide  the  product  by  the  same  number,  the  quotient 
will  be  the  first  number.  If  the  division  be  performed 
first,  then  the  multiplication,  the  result  will  still  be  the 
first  number. 

16 


16  ALGEBRAIC   ARITHMETIC 

Thus,  a  X  6  -J-  6  =  a  ^  6  X  6  =  a,  (2) 

ab     a      L 
or  —  =  -  X  6  =  a. 

0         0 

On  account  of  this  inverse  relation,  every  fact  in 
addition  gives  one  or  more  corresponding  facts  in  sub- 
traction, and  similarly  for  multiplication  and  division. 

Thus,  since  5  +  7  =  12,  it  follows  that 

(7  _|.  5)  _  5  =  7^  or  12  -  5  =  7, 
and  that  (5  +  7)  -  7  =  5,  or  12  -  7  =  5; 
and  since  6  x  8  =  48,  it  follows  that 

(6  X  8)  --  8  =  6,  or  48  -^  8  =  6, 
and  that        (8  x  6)  -^  6  =  8,  or  48  -r-  6  =  8. 

ADDITION 

10.  If  we  have  a  group  of  a  things  and  a  second  group 
of  h  things,  and  if  we  form  a  single  group  from  these 
by  putting  the  two  groups  together,  we  shall  have  as 
many  things  in  the  single  group  thus  formed  as  there 
were  at  first  in  both  the  groups.  It  is  clear  that  this 
will  be  true  whether  we  put  the  first  group  with  the 
second  or  the  second  with  the  first;  that  is,  the  sum 
of  the  things  is  not  changed  by  the  way  in  which  they 
are  brought  together. 

This  fact  or  law  is  expressed  symbolically  thus : 

a  +  b=:b  +  a\  (1) 

and  briefly  in  words  thus : 

Additions  may  he  performed  in  any  order. 

Thus,  for  any  three  numbers  a,  6,  and  c, 
a+6-fc=a+cH-6=64-a  +  c=a+(6H-c)  =  (6-|-a)-|-c,  etc. 


SUBTRACTION  17 

11.  If  I  take  a  marbles  from  a  box  twice,  and  after- 
wards take  the  same  number  of  marbles  3  times,  I  have 
in  all  taken  a  marbles  (2  -f  3)  times,  or  5  times ;  which 
makes  5  a  marbles. 

Thus  the  number  I  take  the  first  time  is 

a -{-a,  or  2  a, 
and  the  second  time,  a-\-  a-\-a,  or  3  a. 

Hence  I  take  in  all,  a-\-a-\-a-\-a-{-ay  or  5a. 
Hence  2a-f3a  =  (2  +  3)a  =  5a. 

If  I  take  a  things  m  times,  and  again  n  times,  I  take 
in  all  a  things  (m  -f-  n)  times,  or  {m  +  n)  a  things. 

Hence  ma  -{-  na  =  {m  -\-  n)  a.  (1) 

From  the  formula  we  have  the  rule :  To  add  terms 
having  a  common  factor,  write  the  common  factor  with  a 
coefficient  equal  to  the  sum  of  the  coefficients  of  the  terms 
added. 

Ex.  1.   3a6  +  5a6  +  a?>  =  9a6. 

Ex.  2.   4ac-f-66c=(4a  +  66)c. 

Exercise.  Show  that  the  above  equations  are  true 
when  a  =  l,  6  =  2,  c  =  3;  when  a  =  4,  6  =  3,  c  =  2; 
when  a  =  h  =  c  =  ^. 

Are  they  true  for  all  values  of  the  letters  ? 

SUBTRACTION 

12.  If  there  are  a  apples  in  one  basket  and  h  in 
another,  and  I  take  away  c  of  them,  the  number  remain- 
ing will  be  a  -f  6  —  c.  This  result  does  not  show  whether 
I  take  the  apples  partly  from  each  basket  or  all  from 


18  ALGEBRAIC   ARITHMETIC 

one.  It  merely  indicates  that  the  whole  number  of 
apples,  a  -h  6,  has  been  diminished  by  c. 

If  I  take  them  all  from  the  first  basket,  the  number 
remaining  in  it  will  be  a  —  c,  and  the  whole  number  of 
apples  remaining  will  be  a  —  c  +  6.  Similarly,  if  I  take 
them  all  from  the  second  basket,  the  whole  number  re- 
maining will  be  a-\-(b  —  c)j  or  6  —  c  +  a. 

The  whole  number  remaining  will  be  the  same  which- 
ever way  I  take  the  c  apples ;  hence 

a-f-6  —  <?  =  a  —  cH-6  =  6  —  c-fa.  (1) 

From  this  we  have  the  law :  Subtractions  may  be  per- 
formed in  any  order. 

Note.  This  law  is  limited  to  the  case  where  the  minuend  is  at 
least  as  large  as  the  subtrahend. 

13.  From  articles  10  and  12  it  is  clear  that 

a+(6  +  c)=fl  +  6  +  (?,  (1) 

(H-(6-c)=a-f  6-(?.  (2) 

Hence  a  poA'enthesis  which  is  preceded  by  the  sign  of 
addition  may  be  removed  from  an  expression  without  affect- 
ing its  value. 

14.  The  expression  a—(b-{-c)  means  that  from  a  we 
are  to  subtract  the  sum  of  b  and  c.  We  shall  evidently 
obtain  the  same  result  by  first  subtracting  b  from  a,  then 
subtracting  c  from  the  remainder  •,  hence 

a-(b-\-c)=a-b-c.  (1) 

The  expression  a  —  (b  —  c)  means  that  from  a  we  are 
to  subtract  the  difference  between  b  and  c ;  hence  if  we 


SUBTRACTION  19 

subtract  b  from  a,  we  subtract  c  units  too  much.  Hence, 
to  obtain  the  correct  result,  we  must  add  c  to  the  re- 
mainder.    That  is, 

a-(b-c)=a-b  +  c.  (2) 

From  (1)  and  (2)  it  follows  that  a  parenthesis  luhich 
IS  preceded  by  the  sign  of  subtraction  may  be  removed 
from  an  expression  if  all  the  -f-  signs  within  the  paren- 
thesis be  changed  to  —  signs,  and  all  the  —  signs  to  -\- 
signs. 

Ex.   a-(c-2&  +  5)=a-c  +  26-5. 

EXAMPLES  4 

If  a  =  6,  6  =  5,  c  =  4,  d  =  2,  and  e  =  1,  find  the 
values  of  the  following  expressions  (1)  by  substituting 
in  the  given  expressions,  then  performing  the  indicated 
operations,  (2)  by  removing  parentheses,  combining  like 
terms,  then  substituting  the  values  of  the  letters.  The 
results  should  agree. 

1.  3ad+(2c-ae).  8.   3a-(26-a  +  5). 

2.  bde-{a-2d).  9.   Zad-{ad-b)-\-2b-c. 

4.  7cd+(8a-5cd  +  9).       11.   20e-[3a-(26  +  a)]. 

5.  5a&4-[6c-(3c-5e)].    12.   2 6c  - (a6  -  6c)  +  2 a6. 

6    ^—(p  —  c).  13    a4-(2c  — ci)      e 

d  '  c  a 


20  ALGEBRAIC   ARITHMETIC 

MULTIPLICATION 

15.  For  the  case  where  thfe  multiplier  is  an  integer, 
multiplication  is  defined  as  the  process  of  taking  one 
number  as  many  times  as  there  are  units  in  another 
number. 

Thus  3x5  =  5  +  5  +  5 

(the  5  being  taken  as  many  times  as  there  are  units  in 
the  multiplier  3), 
and  4a  =  a  +  a  +  a  +  a. 

This  definition  fails  when  the  multiplier  is  a  mixed 
number  or  a  fraction,  for  we  cannot  take  anything  a 
fraction  of  a  time.  A  fractional  multiplier  does  not 
indicate  how  many  times  the  multiplicand  is  to  be  taken, 
but  what  part  of  it. 

Thus  f  X  a  means  that  3  of  the  4  equal  parts  of  a 
are  to  be  taken.  The  multiplier  itself  is  3  of  the  4  equal 
parts  of  unity.  Hence  we  have  the  following  definition 
of  multiplication,  which  holds  for  any  value  of  the  mul- 
tiplier : 

Multiplication  is  the  process  of  doing  to  the  multipli- 
cand what  was  done  to  unity  to  obtain  the  multiplier. 

Numerical  examples  will  make  the  meaning  clear. 

Multiply  10  by  |. 

To  obtain  the  multiplier,  1  was  divided  into  5  equal 
parts,  and  two  of  these  parts  were  taken.  Hence  divide 
10  into  5  equal  parts  and  take  two  of  them.  The  result 
is  4. 

Multiply  5  by  3. 

The  multiplier  is  three  ones  (1  + 1  + 1). 

Hence  the  product  is  three  fives  {b-^-b-^-  6). 


MULTIPLICATION  21 

16.  The  number  of  dots  in  the  figure,  counted  by  rows, 

is  three  fives ;  counted  by  columns,  it  is  five 
threes.    The  number  of  dots  must  he  the  same 
•  •  •  •  •     whichever  way  they  are  counted;   hence 
3x5  =  5x3. 
The  same  reasoning  holds  for  the  product  of  any  two 
integers.     Thus  if  there  are  a  rows  of  6  dots  each,  the 
whole  number  of  dots,  counted  by  rows,  will  be  ah ;  and 
if  counted  by  columns  it  will  be  ha. 
Hence  ab  =  ba.  (1) 

This  is  a  law  of  multiplication,  which,  expressed  in 
words,  is :  The  order  in  which  factors  are  taken  does  not 
affect  the  value  of  the  product ;  or,  more  briefly : 
The  factors  of  a  product  may  he  taken  in  any  order. 

Note.  It  should  be  observed  that,  in  the  proof  of  this  law,  we 
have  assumed  that  a  and  h  are  integers.  The  law  holds  good  when 
either  or  both  the  factors  are  fractions ;  but  the  proof  is  different 
and  rather  more  difficult. 

The  law  holds  for  any  number  of  factors;  thus  for 
three  factors  a,  h,  and  c, 

ahc  =  ach  =  cha  =  a(hc),  etc.  (2) 

17.  It  follows  directly  from  the  law  stated  in  the  pre- 
ceding article  that  a  numher  is  multiplied  hy  multiplying 
any  one  of  its  factors. 

Thus,  if  we  wish  to  multiply  the  number  ahc  by  any 
number  x,  we  know  from  this  law  that 

(abc)  X  =  (ax)  bc  =  a  (bx)  c  =  ab  (ex).  (1) 

Exercise.  Show  that  30  is  multiplied  by  7  by  multi- 
plying any  one  of  its  prime  factors  by  7. 


22  ALGEBRAIC   ARITHMETIC 

18.  To  multiply  24  by  2  we  multiply  the  4  units  by 
2  and  the  2  tens  by  2 ;  that  is, 

2  X  24  =  2(20  +  4)=  2  X  20  +  2  X  4. 

We  may  separate  the  multiplicand  into  parts  in  any 
way,  and  multiply  it  by  multiplying  each  of  those  parts. 
For  example, 

2x24  =  2(12 +  7 +  5)=  2x12  4-2x7 +  2x5. 

The  same  fact  is  true  of  any  number. 

Thus,  o  (/fi  4-  /i)  =  a/n  +  an.  (1) 

Hence  the  law :  A  number  is  multiplied  by  multiplying 
each  of  its  parts  (terms). 

Note.  The  parts  of  a  number  are  not  its  factors.  A  number 
is  produced  from  its  parts  by  addition,  not  by  multiplication. 

Ex.1.   5(3a6  +  2)=15a6  +  10. 

Ex.2.   2a(b  +  cd)=2ab-\-2acd. 

Ex.  3.   3b(cd-{-S^-\-^e)  =  3bcd  +  9a-^2e. 

Exercise.  Show  that  these  equations  are  true  when 
a  =  1,  6  =  2,  c  =  5,  and  d  =  3 ;  also  when  a  =  6,  6  =  1, 
c  =  2,  and  d  =  3.  Are  they  true  for  other  values  of  the 
letters  ? 

19.  Not  only  may  the  multiplicand  be  separated  into 
parts,  but  the  multiplier  may  be  also.  This  is  done  in 
finding  a  numerical  product  when  the  multiplier  consists 
of  more  than  one  figure. 


MULTIPLICATION  23 


Thus  the  operation  of  finding 

42  X  35,  when  expressed 

ily,  IS 

35  = 

30  +  5 

42  = 

40  +  2 

2x    5  = 

10 

2x30  = 

60 

40  X    5  = 

200 

40x30  = 

1200 

42  X  35  =  1470 

If  we  should  separate  the  factors  into  parts  in  any  way, 
and  should  multiply  each  part  of  the  multiplicand  by 
each  part  of  the  multiplier,  the  sum  of  these  partial 
products  would  be  the  product  of  the  factors. 

Exercise.  Find  the  product  of  35  and  42  after  separ 
rating  the  factors  into  the  parts :   35  =  12  +  20  +  3,  and 

42  =  20  +  22. 

Note.  The  sign  of  multiplication  is  omitted  between  paren- 
theses, and  between  a  parenthesis  and  a  factor. 

Ex.1.    (a  +  6)(c  +  d)=a(c  +  (r)+6(c  +  c?) 
=  ac  +  ac^  +  6c  +  bd. 

Ex.2.   2a(6  +  4c)=2a6  +  8ac. 

Ex.  3.    (3a  +  6)(c  +  5e)=3a(c  +  5e)+&(c  +  5e) 
=  3  ac  +  15  ae  +  6c  +  5  be. 

Exercise.  Show  that  the  above  results  are  true  when 
a  =  1,  6  =  c  =  2,  c?  =  4,  and  e  =  3. 

Give  the  letters  a  different  set  of  values,  and  show  that 
the  results  are  true  for  those  values. 


24  ALGEBRAIC   ARITHMETIC 

20.  The  expression  4  (8  —  3)  means  that  the  difference 
between  8  and  3  is  to  be  taken  4  times. 

Hence  4(8  -  3)=  4  x  5  =  20. 

The  result  can  be  found  differently  as  follows :  If  we 
take  4x8,  every  time  we  have  taken  8  instead  of  5  we 
have  taken  3  too  many.  Hence  we  have  taken  3  too 
many  4  times,  or  4  x  3  too  many  in  all ;  and  the  result 
will  be  correct  if  we  subtract  that  number. 

Hence  4(8  -  3)=  4  x  8  -  4  x  3. 

The  same  reasoning  holds  for  any  numbers ;  hence,  in 
general, 

a(b  —  c)=ab  —  ac.  (1) 

Hence  the  law  given  in  Art.  18  may  be  extended  so  as 
to  read :  An  expression  is  multiplied  by  multiplying  each 
of  its  terms,  whether  they  are  to  be  added  or  subtracted. 

21.*  To  multiply  (c  —  c^)  by  (a  —  b),  first  take  (c  —  d) 
a  times,  which  gives  a(c  —  d),  or  ac  —  ad.  This  result  is 
too  large,  for  in  taking  the  multiplicand  a  times  instead 
of  (a  —  b)  times,  we  have  taken  it  b  times  too  many. 
Hence  we  must  subtract  6  (c  —  d),  or  (be  —  bd). 

Hence  {a  —  b){c  —  d)=  a{c  —  d)—  b{c  —  d) 
=  a{c  —  d)  —  (bc  —  bd) 
=  ac  —  ad  —  bc-{-  bd. 

From  this  example  we  may  deduce  the  following  rule 
for  the  multiplication  of  algebraic  quantities:  Multiply 
each  term  of  the  multiplicand  by  each  term  of  the  multiplier. 

»  This  article  may  be  omitted  at  the  discretion  of  the  teacher. 


MULTIPLICATION  25 

When  the  two  terms  of  a  product  have  both  +  or  both  — 
before  them,  put  -f-  before  their  product;  when  one  has  + 
and  the  other  —,  put  —  before  their  product.  In  using  the 
first  terms  of  the  expressions,  which  have  no  sign,  apply  the 
rule  as  if  they  had  the  -f-  sign. 

The  rule  for  the  signs  may  be  briefly  stated:  Like 
signs  give  -\-  and  unlike  signs  give  — . 

Ex.     {2a-b)(Sa-2b)=2a(Sa-2b)-b(Sa-2b) 
=  6aa-4:ab-Sab-{-2bb 
=  6aa-7ab-\-2bb. 

Exercise.  Show  that  this  result  is  true  when  a  =  3 
and  6  =  2;   when  a  =  6  and  6  =  5. 

22.  When  a  factor  is  to  be  taken  more  than  once  in  a 
product,  instead  of  repeating  the  factor  the  required 
number  of  times,  it  is  written  only  once  with  a  small 
figure  to  the  right  and  a  little  above  it.  This  figure 
shows  how  many  times  the  factor  is  to  be  repeated,  and 
is  called  an  exponent. 

Thus,  52  =  5  X  5,  23  =  2  X  2  X  2.  The  answer  to  the 
example  at  the  end  of  the  last  article  would  be  written 

6a^-7ab-h2b^ 

It  should  be  noticed  that  a  coefficient  and  an  exponent 
have  very  different  meanings. 

Thus  3x6  =  6+6  +  6  =  18; 

63  =  6x6x6  =  216. 
4:a  =  a-\-a-\-a  +  a; 


26  ALGEBRAIC   ARITHMETIC 

Ex.  1.   2ax5a^  =  2x5aaa  =  10a\ 
Ex.  2.   3a(a-4:b)=3a'-12ab. 
Ex.  3.    (a-\-by=(a  +  b)(a  +  b)=a(a-{-b)+b(a  +  b) 
=  a:'-{-ab-\-ab  +  b^  =  a^-{-2ab-{-  Jy". 

Exercise.  Show  that  the  result  of  Ex.  3  is  true  when 
a  =  20  and  6  =  4;  when  a  =  40  and  6  =  3;  when  a  =  4 
and  6  =  3. 

EXAMPLES  5 
Multiply : 

1.  2a  by  4a.  7.  a  +  6  by  3. 

2.  3  a  by  5  cd  8.  3  a  — 6  by  5. 

3.  ^o?'  by  a6.  9.  o?-\-a  by  a. 

4.  2a62  by  10a26.  10.  2a2-a  by  a\ 

5.  4a62c  by  2ca?.  11.  a2  +  2a-2  by  3a. 

6.  Ice  by  ^o?y.  12.  6c-|-ca  +  «6  by  a6c. 

If  a  =  3,  6  =  2,  c  =  1,  and  d==5,  find  the  numerical 
values  of : 

13.  2a^c.  19.  a(d-c-b). 

14.  5a62d  20.  Sb^id'-ac). 

15.  a^b^c'd^  21.  (a -\- b)(d  -  c). 

16.  6a6''  +  3c*.  22.  (a2  + 2  6)(a  -  6). 

17.  5a^c-b\  23.  (a  +  6)*. 

18.  3(a«  +  6).  24.  (2  6 +  <«)'. 


DIVISION  27 

Simplify  the  following  by  removing  parentheses  and 
combining  like  terms : 

25.  2(a-b)+S(a-\-b).         28.    7  a(b  -  c)-2b(a- c). 

26.  Sa(b-\-c)  —  (ab-\-2ac).   29.   |(6  -  2c)- |(c- 2&).\/' 

27.  c(a  +  b)-c(a-b).  30.   2[3a&  -  4a(c  -  2  6)]. 

Find: 

31.  (x  +  yy.  37.*  (a -by. 

32.  (2a-^by.  38.  (2a;-32/)2. 

33.  (a2  +  c)2.  39.  (a^  -  cy. 

34.  (a  +  6)(3a  +  26).  40.  (a-b)(2a-b). 

35.  (a-6)(2a  +  5&).  41.  (2a  -  ft^^^g^  _  252^. 

36.  (a  +  b)(a-2b).  42.  (a  +  6)^. 

DIVISION 

23.  We  have  already  referred  to  division  as  the  inverse 
of  multiplication  (Art.  9).  It  is  the  process  by  which, 
when  the  product  of  two  factors  (the  dividend)  and  one 
of  the  factors  (the  divisor)  are  given,  the  other  factor 
(the  quotient)  is  found. 

In  consequence  of  this  relation  between  the  two  pro- 
cesses, it  is  easy  to  derive  the  laws  and  rules  of  division 
from  the  corresponding  laws  and  rules  of  multiplication. 
We  shall  proceed  to  do  this. 

24.  Since  a  number  is  multiplied  by  multiplying  any 
one  of  its  factors  (Art.  17),  it  follows  that  a  number  is 
divided  by  dividing  any  one  of  its  factors. 

Thus    (abc)-^d=-^xbc  =  ax^xc  =  abx^-         (1) 
d  a  a 

*  This  and  the  following  are  to  be  taken  or  omitted  with  Art.  21. 


28  ALGEBRAIC   ARITHMETIC 

It  will  be  seen  from  this  that  the  result  is  the  same 
whether  the  division  is  performed  before  or  after  any  or 
all  of  the  multiplications  ;  hence : 

Divisions  may  he  performed  in  any  order. 

Ex.  1.  Show  that  336  is  divided  by  2  by  dividing  any 
one  of  its  factors  4,  6,  and  14,  by  2. 

Ex.  2.  Show  that  42  is  divided  by  3  by  dividing  any 
one  of  its  prime  factors  by  3. 

25.  Since  a  number  is  multiplied  by  multiplying  each 
of  its  parts  (Art.  18),  it  follows  that  a  number  is  divided 
by  dividing  each  of  its  parts. 

.Thus  (6  +  c)-^/7i=-  +  -.  (1) 

^         ^  mm  ^  ^ 

This  law  is  employed  in  every  numerical  example  in 
division.  Thus  the  steps  in  the  process  of  dividing  762 
by  3  are  as  follows  : 

762  =  600  4- 150  + 12, 
3)762  =  :&oo+^f^  +  ¥ 
=  200  +  50  +  4 
=  254. 

If  the  method  of  solution  seems  unfamiliar,  it  is 
because  we  are  accustomed  to  perform  the  separate  steps 
mentally,  and  to  put  down  only  the  result. 

26.  It  follows  from  Art.  20,  that  an  expression  is 
divided  by  dividing  each  of  its  terms,  whether  they  are  to  be 
added  or  subtracted. 

Thus  {a-b)-^c  =  ^-^'  (1) 


PROBLEMS  29 

EXAMPLES  6 
Divide : 

1.  15a  by  5a.  7.  12a^  +  15a^b  by  3a. 

2.  12a2by4a.  8.  18 a^ft  -  12 ac  by  6 a. 

3.  Sab  by  26.  9.  5x^y  —  xy^  by  7 xy. 

4.  6xy  by  2a;2/2.  lo.  a^b-^-Sbcd  by  3a. 

5.  30  a^dc  by  3 ac.  11.  (a  +  &)^  by  a  +  &. 

6.  a'b'cd'  by  b'd.  12.  15  a^d^  +  5  a'b^  -  3  a*b  by  5  a^ft. 

PROBLEMS 
Note.     See  Art.  8  for  directions. 

13.  Three  boys,  counting  their  money,  found  they  had 
190  cents.  The  second  had  twice  as  many  cents  as  the 
first,  and  the  third  as  many  as  both  the  others,  plus  4 
cents.     How  many  cents  had  each  ? 

14.  A  cistern  filled  with  water  has  two  faucets,  one  of 
which  will  empty  it  in  5  hr.,  the  other  in  20  hr.  How 
long  will  it  take  both  to  empty  it  ? 

15.  If  12  be  added  to  the  half  of  a  certain  number, 
the  sum  will  be  20.     Find  the  number. 

16.  A  farmer  divided  52  apples  among  3  boys  in  such 
a  manner  that  B  had  ^  as  many  as  A,  and  C  had  2  less 
than  }  as  many  as  A.     How  many  had  each  ? 

17.  The  whole  number  of  hands  employed  in  a  factory 
is  1000.  There  are  twice  as  many  boys  as  men,  and  11 
times  as  many  women  as  boys.  How  many  of  each  are 
there  ? 


30  ALGEBRAIC   ARITHMETIC 

18.  A  and  B  invest  equal  amounts  in  trade.  A  gains 
$  1260,  and  B  loses  ^  870 ;  A's  money  is  now  double  B's. 
What  sum  did  each  invest  ? 

19.  Divide  100  into  two  parts  such  that  twice  one  part 
is  equal  to  3  times  the  other. 

20.  The  sum  of  two  numbers  is  36,  and  their  differ- 
ence is  half  the  greater.     Find  them. 

21.  A  man  of  40  has  a  son  10  yr.  old.  In  how  many 
years  will  the  father  be  3  times  as  old  as  the  son  ? 

22.  A  father's  age  is  3  times  that  of  his  son,  and  in 
10  yr.  it  will  be  twice  as  great.     How  old  are  they  ? 

23.  A  has  $15  more  than  B,  B  has  $5  less  than  C, 
and  they  have  $  65  in  all.     How  much  has  each  ? 

24.  In  a  regiment  containing  1200  men,  there  were  3 
times  as  many  cavalry  as  artillery  less  20,  and  92  more 
infantry  than  cavalry.     How  many  of  each  ? 

25.  What  are  the  ages  of  three  brothers,  whose  united 
ages  are  48  years,  and  their  birthdays  2  years  apart  ? 

26.  The  difference  of  the  squares  of  two  consecutive 
numbers  is  15.     What  are  the  numbers  ? 

27.  At  the  time  of  marriage,  a  man  was  twice  as  old 
as  his  wife ;  but  18  years  later  his  age  was  |  times  hers. 
Required  their  ages  on  the  wedding  day. 


CHAPTER   III 

PERCENTAGE  AND  ITS  APPLICATIONS 

PERCENTAGE 

27.  Three  closely  related  operations  are  frequently- 
employed  in  commercial,  or  business  arithmetic;  namely: 

I.  To  find  a  certain  part  of  a  number. 
II.  To  find  what  part  one  number  is  of  another. 
III.  To  find  a  number  when  a  certain  part  of  it  is  given. 

Ex.  1.   What  is  I  of  50? 

What  part  of  50  is  30  ? 

What  is  the  number  of  which  30  is  f  ? 

Ex.  2.  A  man  had  75  sheep,  and  he  sold  f  of  them. 
How  many  did  he  sell  ? 

A  man  had  75  sheep,  and  he  sold  50  of  them.  What 
part  of  his  sheep  did  he  sell  ? 

A  man  sold  50  sheep,  which  was  J  of  all  he  had.  How 
many  had  he  at  first  ? 

It  will  now  be  seen  what  is  meant  by  saying  that  these 
operations  are  closely  related.  The  three  questions  in 
each  of  the  examples  involve  the  same  three  numbers, 
of  which  two  are  given  and  the  third  required ;  and  any 
one  of  the  three  can  be  found  if  the  other  two  are  given. 

28.  It  is  customary  in  business  to  express  the  frac- 
tion that  one  number  is  of  another  in  hundredths,  even 

31 


32  ALGEBRAIC   ARITHJVIETIC 

when  the  fraction   can    be    readily   reduced   to   lower 
terms. 

Thus4is^of  8;3is^of30;  lis^of40.. 

In  stating  problems,  the  denominator  100  is  omitted, 
and  the  phrase  per  centy  which  means  hundredths,  is  used 
instead. 

The  sign  %  means  per  cent. 

The  following  expressions  exhibit  the  different  ways 
of  denoting  a  fractional  part : 

i=^  =  .50   =50  per  cent  =  50%. 
2     100  ^  ^ 

i  =  ^  =  .125  =  12|.  per  cent  =  12^%. 

1         i 
200  ==  100  "  .005  =  \  per  cent  =  i%. 

5  =  1?^  =  1.25  =  125  per  cent  =  125%. 

1  =  152  =  1.00  =  100  per  cent  =  100%. 

Note.  It  should  be  remembered  that  100%  of  a  number  is 
once  the  number,  or  the  number  itself. 

EXAMPLES  7  (Oral) 

Name  the  corresponding  fractions  in  lowest  terms : 

2%  15%  40%  80%  100%  \% 

4%  20%  45%  85%  120%  i% 

6%  25%  60%  90%  125%  y\% 

10%  35%  75%  95%         175%         12^% 


PERCENTAGE  33 

EXAMPLES  8 


Express  as  fractions  in  the  lowest  terms.    The  results 
should  be  memorized : 

6i% 

8W 
12i% 

16|% 
33J% 

62i% 
66f% 
83i% 

87i% 
2*% 

BXAMPT.KS  9 

Express  as 

per  cent. 

Memor: 

ize  the  first  four  columns : 

* 

TJHF 

A 
A 

« 

29.  In  computations,  per  cent  is  expressed  as  a  com- 
mon fraction  (in  lowest  terms)  or  as  a  decimal,  according 
as  the  one  form  or  the  other  is  the  more  convenient. 

Ex.  1.   What  is  20%  of  85?  (Case  I,  Art.  27.) 

20%  of  a  number  is  ^  of  it ;  and  \  of  85  is  17. 

Ex.  2.   What  per  cent  of  30  is  18  ?  (Case  11.) 

18  is  H  of  30 ;  and  ^  is  f,  or  60%. 

Ex.  3.   8  is  48%  of  what  number  ?  (Case  III.) 

Since  48  is  8%  of  the  number,  1%  of  it  is  |-  of  48, 

or  6;  and  100%  of  the  number  is  100  x  6,  or  600.     Or, 
Since  8%,  or  -^,  of  the  number  is  48,  -^  of  it  is  ^  of 

48,  or  24 ;  and  |f  of  the  number  is  25  x  24,  or  600. 

Ex.  4.  What  number  diminished  by  5%  of  itself  is 
38  ?  (Case  III.) 

100%  -  5%,  or  95%,  of  the  number  is  38.  95%  =  |^. 
Hence  the  number  is  20  times  ^  of  38,  or  40. 


84  ALGEBRAIC   ARITHMETIC 

EXAMPLES  10  (Oral) 
Find 

1.  5%  of  120.  5.  8%  of  300  sheep. 

2.  121%  of  72.  6.  6i%  of  f  320. 

3.  25%  of  96.  7.  81%  of  24  men. 

4.  33i%  of  66.  8.  75%  of  300  bu. 

What  per  cent  of 

9.   40  is  15?  13.  72  rd.  are  18  rd.? 

10.  12^  is  2|?  14.  1  T.  are  2  cwt.? 

11.  1  da.  are  16  hr.?  15.  1  gal.  are  3  pt.? 

12.  lib.  are  2oz.?  16.  ^480  are  ^24? 

What  is  the  number  of  which 

17.  30  is  20%?  21.  96  is  133J%? 

18.  84  is  12%?  22.  55  is  125%? 

19.  5  is  i%?  23.  72  is  66|%? 

20.  16  is  32%?  24.  15  is  16f%? 

25.  A  farmer  had  150  sheep,  and  sold  16f  %  of  them. 
How  many  did  he  sell  ? 

26.  A  boy  increasing  his  money  by  25%  of  itself  has 
$  1.     What  had  he  at  first  ? 


27.  A  gi'ocer  bought  150  dozen  eggs,  and  found 
of  them  bad  or  broken.     How  many  were  salable  ? 

28.  What  number  increased  by  8J%  of  itself  is  130  ? 

29.  A  clerk  has  his  salary  increased  12^%,  and  he 
then  gets  $  18  per  week.  What  was  his  salary  before 
the  increase? 


PERCENTAGE  36 

30.  A  man  sold  a  horse  for  $100  at  20%  above  cost. 

Find  the  cost. 

31.  In  a  school  of  75  pupils  3  were  absent.  What  per 
cent  was  absent? 

32.  John  has  36  cents,  which  is  37|-%  of  what  his 
brother  has.     How  much  has  his  brother? 

33.  A  clerk  spends  88%  of  his  salary  and  saves  $144. 
What  is  his  salary  ? 

34.  37^%  of  a  stock  of  goods  valued  at  $1200  was 
destroyed  by  fire.     What  was  the  loss? 

30.  Percentage  includes  all  operations  in  which  a  per 
cent  of  a  number  is  given  or  required. 

The  number  of  which  the  per  cent  is  taken  is  called 
the  base. 

The  per  cent,  when  expressed  decimally  or  as  a  com- 
mon fraction,  is  usually  called  the  rate  per  cent,  or  simply 
the  rate. 

The  percentage  is  the  result  obtained  by  taking  a  cer- 
tain per  cent  of  the  base.  It  is,  therefore,  a  product^  of 
which  the  factors  are  the  base  and  the  rate. 

Note.  It  will  be  seen,  from  the  two  definitions  of  percentage, 
that  the  word  is  used  (1)  to  name  a  class  of  operations,  (2)  to 
name  the  result  of  an  operation. 

31.  If  we  use  the  initial  letters  of  the  words  base,  rate, 
and  percentage  to  denote  the  numbers  called  by  these 
names,  we  can  easily  express  algebraically  the  relations 
that  these  numbers  bear  to  one  another.  The  equations 
expressing  these  relations  are  percentage  formulas. 


36  ALGEBRAIC   ARITHMETIC 

From  the  definition  of  percentage,  we  know  that 

p  =  bt:  (1) 

Ex.  1.  A  man  invests  $1265,  and  gains  12%  on  his 
investment.     How  much  does  he  gain  ? 

6  =  $1265,  r=.12,   p^? 
$1265 

.\2 

$151.80=  p. 

Divide  the  members  of  (1)  by  h ;  then 

f  =  .or.  =  |.  (2) 

Here  we  have  given  a  product  (percentage)  and  one  of 
the  factors  (base)  to  find  the  other  factor  (rate  per  cent). 

Ex.  2.  A  merchant  owes  $15,120,"  of  which  he  can 
pay  only  $9828.  What  per  cent  of  his  debts  can  he 
pay? 

p  =  $9828,   6  =  $15120,  r  =  ? 

$9828  ^  gg^gg^ 
$15120  /"^ 

Divide  the  members  of  (1)  by  r ;  then 


(3) 


Ex.  3.   What  number  increased  by  18%  of  itself  is 
2950  ? 
2950  is  100%  +  18%  or  118%  of  the  number;  hence 
ly  =  2950,  r  =  1.18,  6  =  ? 

6  =  ????  =  2500. 
1.18 


PERCENTAGE  37 

Note.  In  some  problems,  as  in  this  one,  one  of  the  given  num- 
bers is  not  directly  stated  ;  but  must  be  found  from  the  conditions 
of  the  problem  before  the  formula  can  be  applied. 

32.  Percentage  Formulas.  The  three  cases  of  percent- 
age and  their  formulas  are  : 

Case  I.    To  find  a  given  per  cent  of  a  number. 

p  =  br.  (1) 

Case  II.    To  find  what  per  cent  one  number  is  of  another. 

-f-  (2) 

Case  III.  To  find  a  number  when  a  certain  per  cent  of 
it  is  gvven. 

*=^-  (3) 

EXAMPLES  11 

1.   A  merchant  failing  was  able  to  pay  his  creditors  but 
^0.     He  owes  A  $  3500,  B  $  1200,  C  $  1134,  D  $  650. 


What  will  each  receive  ? 

2.  A  person  whose  annual  income  is  f  450  pays  $  125 
for  board,  $  140  for  clothing,  f  25  for  books,  and  $  30  for 
sundries.  What  per  cent  of  his  income  is  each  item,  and 
what  per  cent  remains  ? 

3.  The  deaths  in  a  certain  city,  during  the  year,  are 
980,  which  is  3^%  of  the  population.  What  is  the  popu- 
lation ? 

4.  Sold  cloth  for  $3.50  per  yard,  which  was  70%  of 
its  cost.     What  was  the  cost  per  yard  ? 


38  ALGEBRAIC   ARITHMETIC 

5.  A  merchant  failing  owes  $3500;  his  property  is 
valued  at  $  2100.  What  per  cent  of  his  indebtedness  can 
he  pay  ? 

6.  A  shepherd  lost  12%  of  a  flock  of  sheep  by  disease, 
and  then  had  2200.  How  many  were  in  the  flock  at 
first? 

7.  Sold  a  house  and  lot,  which  cost  me  $  1450.75,  at  a 
gain  of  15%.     What  was  the  gain  ? 

8.  A  man  spent  in  one  year  $  2150,  which  was  5|% 
of  what  he  had.     How  much  had  he  ? 

9.  A  man  having  $5800  worth  of  hay  lost  $870 
worth  by  fire.  What  per  cent  of  the  whole  was  the  part 
lost  ? 

10.  A  tailor,  after  using  75%  of  a  piece  of  cloth,  had 
9 J  yards  left.     How  many  yards  were  in  the  whole  piece  ? 

11.  A  man  drew  25%  of  his  bank  deposits,  and  spent 
33  J  %  of  the  money  thus  drawn  in  the  purchase  of  a 
horse  worth  $250.  How  much  money  had  he  in  the 
bank  at  first  ? 

12.  A  man  owning  f  of  a  cotton-mill,  sold  35%  of  his 
share  for  $  24,640.  What  part  of  the  whole  mill  did  he 
still  own,  and  what  was  its  value  ? 

PROFIT  AND  LOSS 

33.  Gains,  losses,  and  selling  price  (S.  P.)  are  always 
reckoned  as  a  per  cent  of  the  cost;  in  other  words,  they 
are  percentages  computed  on  the  cost  as  base. 

The  following  table  shows  what  quantities  are  denoted 


PROFIT   AND   LOSS  39 

by  the  letters  of  the  percentage  formulas  (Art.  32)  in  the 
various  problems  that  occur  in  Profit  and  Loss : 

Table 


b  =  cost, 


'  r  =  rate  of  gain,  p  =  profit.  (1) 

r  =  rate  of  loss,  p  =  loss.  (2) 

r  =  1  +  rate  of  gain, )  ^      ^  p  (3) 


■;1 


r  =  1  —  rate  of  loss,  y         '    '         (4) 


34.  Ex.  1.    A.  man  sells  a  farm  for  $2081.25,  gaining 
11%.     What  did  the  farm  cost  him  ? 

(Case  III,  Art.  32  and  (3)  of  Table.) 

p  =  S.  P.  =  $  2081.25  (=  111  %  of  cost). 

r  =  l.ll,    6  =  cost  =  ? 

6  =  $  2081.25  -^  1.11  =  $  1875.  Ans. 

Ex.  2.   At  what  price  must  goods  that  cost  $3.50  per 
yard  be  sold  to  lose  20%  ? 

(Case  I,  and  (4)  of  Table.) 

The  S.  P.  will  be  100%  -  20%,  or  80%  of  the  cost. 
b  =  $3.50,  r  =  1  -  .2  =  .8,  p  =  S.  P.  =  ? 
jp  =  $  3.50  X  .8  =  $  2.80.  Ans. 

Ex.  3.   Find  the  gain  per  cent  on  a  horse  sold  for  $  72 
at  a  gain  of  $  9.50.     (Case  II,  and  (1)  of  Table.) 

6  =  $  72  -  $  9.50  =  $  62.50  =  cost. 

p  =  $9.50,  r  =  ? 

r  =  $  9.60  -J-  $  62.50  =  .162  =  16^%.  Ana. 


40  ALGEBRAIC   ARITHMETIC 

EXAMPLES  12  (Oral) 

1.  Bought  a  COW  for  $  40,  and  sold  her  for  20%  above 
cost.     What  did  I  receive  for  her  ? 

2.  A  watch  that  cost  $25  was  sold  at  a  loss  of  10%. 
What  was  the  loss,  and  the  selling  price  ? 

3.  A  tailor  bought  cloth  at  $  6  a  3^ard,  and  wished  to 
sell  it  at  a  gain  of  16|%.    At  what  price  must  he  sell  it? 

4.  A  merchant  sells  silk  at  a  profit  of  $  1,50  per  yard, 
which  is  37i%  gain.  What  did  it  cost,  and  what  is  the 
selling  price  ? 

5.  A  watch  was  sold  for  $  34,  at  a  gain  of  6J%.  What 
was  the  cost  ? 

6.  A  dealer  lost  12^%  on  a  reaper  by  selling  it  for 
$  56.     For  what  should  he  have  sold  it  to  gain  12^%  ? 

7.  Sold  melons  for  f  .40  that  cost  $  .30.  What  was 
the  gain  per  cent  ? 

8.  What  per  cent  is  gained  on  an  article  bought  for 
$3  and  sold  for  $5? 

9.  If  corn  selling  for  21  cents  a  bushel  yields  a  profit 
of  50%,  what  did  it  cost  ? 

EXAMPLES  13 

4.  A  man  offers  a  farm,  for  which  he  gave  $  3450, 
for  20%  less  than  its  cost.     What  is  his  price  ? 

Note.  Computations  are  simplified  by  expressing  the  per  cent 
as  a  common  fraction  when  it  is  an  aliquot  part  of  100. 

5.  For  how  much  per  barrel  must  I  sell  flour  costing 
$4.50  per  barrel  to  gain  16|%  ? 

SuGOESTioN.    To  $4. 60  add  ^  of  it. 


PROFIT   AND   LOSS  41 

6.  Sold  a  cargo  of  wheat  for  $  16,000,  at  a  profit  of 
25%.     What  was  the  cost  of  the  cargo  ? 

7.  A  merchant  made  a  profit  of  $  156  by  selling  a 
quantity  of  silks  at  a  gain  of  12%.  What  was  the  cost 
of  the  silks,  and  for  how  much  were  they  sold  ? 

8.  A  merchant  marked  a  piece  of  carpeting  25% 
more  than  it  cost  him,  but,  anxious  to  effect  a  sale,  and 
supposing  he  would  still  gain  5  % ,  sold  it  at  a  discount  of 
20%  from  his  marked  price.     Did  he  gain  or  lose  ? 

Suggestion.  S.  P. = 80  %  of  marked  price  ;  marked  price = 125  % 
of  cost.    S.  P.  =  ?  %  of  cost  ? 

9.  Sold  a  lot  of  books  for  ^480,  and  lost  20%,.  For 
what  should  I  have  sold  them  to  gain  20%  ? 

10.  A  man  bought  a  pair  of  horses  for  $450,  which 
was  25%  less  than  their  real  value,  and  sold  them  for 
25%  more  than  their  real  value.     What  was  his  gain? 

Suggestion.     The  real  value  is  the  base  in  both  operations. 

11.  A  merchant  pays  $6840  for  a  stock  of  goods,  and 
sells  them  at  an  advance  of  26|^%  on  the  purchase  price. 
After  deducting  $  375  for  expenses,  what  is  his  gain  ? 

12.  A  dealer  bought  108  bbl.  of  apples  at  $4.62^,  and 
sold  them  so  as  to  gain  $  114.88^.  What  was  his  gain 
per  cent  ? 

13.  My  goods  are  marked  to  sell  at  retail  at  40% 
above  cost.  I  furnish  my  wholesale  customers  at  12% 
discount  from  the  retail  price.  What  per  cent  profit  do 
I  make  on  goods  sold  at  wholesale  ? 

Suggestion.  88%  of  140%  of  cost,  or  ^^^  x  140%  of  cost - 
wholesale  S.  P. 


42  ALGEBRAIC   ARITHMETIC 

14.  At  what  price  must  shovels  that  cost  $  1.12  each 
be  marked  in  order  to  abate  5%  (of  marked  price),  and 
yet  make  25%  profit  ? 

16.  By  selling  coffee  at  18  cents  per  pound,  I  make  a 
profit  of  20%.  For  how  much  must  I  sell  it  to  make  a 
profit  of  16|%  ? 

16.  Bought  land  at  ^  60  an  acre.  How  much  must  I 
ask  an  acre,  that  I  may  deduct  25%  from  my  asking 
price,  and  still  make  20  %  on  the  cost  ? 

17.  Find  the  loss  per  cent  on  goods  sold  for  $425.98, 
at  a  loss  of  $134.52. 

18.  Sold  goods  for  $  3.50  less  than  cost,  and  lost  14%. 
What  per  cent  should  I  have  gained  by  selling  for  $2.75 
above  cost  ? 

19.  Two  sets  of  furniture  were  sold  for  $35  each. 
On  one  there  was  a  gain  of  16f  %  ;  on  the  other  a  loss  of 
16|%.  Was  there  a  gain  or  a  loss  on  both,  and  how 
much  per  cent  ? 

20.  A  hardware  merchant  bought  three  dozen  agate 
basins  at  the  rate  of  3  for  $  5,  and  sold  them  at  a  gain  of 
$  10  on  the  whole.  What  was  the  average  selling  price 
of  each,  and  what  was  the  gain  per  cent  ? 

21.  I  bought  a  horse  of  Mr.  A.  for  15%  less  than  it 
cost  him,  and  sold  it  for  30%  more  than  I  paid  for  it. 
I  gained  $  15  in  the  transaction.  How  much  did  the 
horse  cost  me  ?  How  much  did  it  cost  Mr.  A.  ?  For 
what  did  I  sell  it  ? 

22.  If  tea,  when  sold  at  a  loss  of  25%,  brings  $1.25 
per  pound,  what  would  be  the  gain  or  loss  per  cent  if 
sold  for  $1.60  per  pound? 


COMMISSION  AND   BROKERAGE  48 


COMMISSION  AND  BROKERAGE 

35.  A  person  who  buys  and  sells  goods  or  lands,  col- 
lects debts,  or  transacts  other  business  of  like  nature  for 
another  person  is  called  a  commission  merchant  or  agent. 

The  pay  received  for  such  services  is  called  commission. 
It  is  usually  a  percentage  on  the  money  paid  for  property 
bought ;  on  the  money  received  for  property  sold ;  on  the 
money  collected. 

A  broker  is  a  person  who  buys  and  sells  stocks,  bonds, 
bills  of  exchange,  etc.,  for  a  commission,  which  is  called 
brokerage. 

The  money  that  remains  from  a  sale  after  the  com- 
mission and  other  expenses  have  been  paid  is  called  the 
net  proceeds. 

Table 

f  am't  of  sale,         \  f ''  =        ^^*®  ^^  com.,  p  =  com.  (1) 

""  I  money  collected,  j  \  r  =  1  —  rate  of  com.,  p  =  proceeds.  (2) 

b  =    am't  of  purchase,       r  =  1  -f  rate  of  com.,  p  =  remit.        (3) 


36.  Ex.  1.     Find  the  commission  on  the  sale  at  auction 
of  a  house  and  the  furniture  for  $9346.80  at  6i%. 

6  =  $  9346.80,  r  =  .06J,  p  =  ? 

p  =  $  9346.80  X  .0625  =  $  584.175  com. 

Ex.  2.   Find  the  net  proceeds  of  the  above  sale. 
b  =  $  9346.80,  r  =  1  -  .06^  =  .93},  p  =  ? 
p  =  $  9346.80  X  .93|  =  $  8762.625  proceeds. 
Or,  from  Ex.  1, 

$9346.80  -  $  584.176  =  $8762.625  proceeds. 


44  ALGEBRAIC    ARITHMETIC 

Ex.  3.  I  send  ^3120  to  a  commission  merchant  to 
buy  flour  at  4%  commission.  Find  cost  of  flour  and 
commission. 

The  remittance  includes  the  investment +4%  of  it; 
hence  is  104%  of  the  investment. 

p  =  $3120,  r  =  1  +  .04  =  1.04,  6  =  ? 
6  =  $  3120  -f- 1.04  =  $  3000.  Ans. 

EXAMPLES  14 

4.  Find  the  commission  on  the  sale  of  a  farm  for 
$13,750,  at  2f%. 

5.  A  commission  merchant  sells  225  bbl.  of  potatoes 
at  $  3.25  per  bbl.,  and  316  bbl.  of  apples  at  |4.50  per  bbl. 
What  is  his  commission  at  4^%  ? 

6.  A  dealer  sends  his  agent  in  Havana  $6720.80, 
with  which  to  purchase  fruits,  after  deducting  his  com- 
mission of  5%.  What  sum  did  the  agent  invest,  and 
what  was  his  commission? 

7.  If  $63  is  paid  for  collecting  a  debt  of  $1260, 
what  is  the  rate  of  commission  ? 

8.  An  architect  charges  |%  for  plans  and  specifica- 
tions, and  1^%  for  superintending  the  construction  of 
a  building  which  cost  $32,000.     What  is  his  fee? 

9.  My  agent  has  purchased  goods  for  me  to  the 
amount  of  $  12.50,  for  which  he  charges  a  commission 
of  If  %.  What  sum  must  I  remit  to  pay  for  goods  and 
commission  ? 

10.  Sent  to  my  agent  in  Cincinnati  $  765  to  purchase 
bacon,  after  deducting  his  commission  of  2%.  What  is 
his  commission,  and  what  does  he  expend  for  bacon  ? 


COMMISSION   AND   BROKERAGE  45 

11.  A  grocer  sends  $2490  to  a  commission  merchant 
to  buy  sugar  at  3|%  commission.  If  he  pays  8  cents 
a  pound  for  the  sugar,  for  what  must  the  grocer  sell  the 
whole  to  gain  16|%  on  the  whole  cost,  and  at  how  much 
per  pound? 

12.  A  collector  collected  rents  at  3%  commission  and 
received  f  87.60  for  his  services.  What  sum  of  money 
did  he  collect? 

13.  I  pay  $275  for  a  lot  and  build  on  it  a  house  cost- 
ing $  1720,  which  my  agent  rents  for  $  25  a  month,  charg- 
ing 5%  commission.  What  per  cent  do  I  make  a  year 
on  the  money  invested  ? 

14.  Find  the  commission  on  the  sale  of  100  bales  of 
cotton,  averaging  480  lb.  to  the  bale,  at  $18  per  cwt., 
the  commission  being  5%. 

15.  An  agent  sells  450  tons  of  hay  at  $  13  a  ton,  com- 
mission 5%,  and  with  the  proceeds  buys  wool  at  22^ 
cents  per  pound,  commission  4%.  What  is  his  whole 
commission,  and  how  many  pounds  of  wool  does  he  buy  ? 

16.  An  agent  in  Boston  received  28,000  lb.  of  cotton, 
which  he  sold  at  $  .12^  per  lb.  He  paid  $  45.86  freight 
and  cartage,  and  after  retaining  his  commission,  he  re- 
mits $3252.89  as  the  net  proceeds  of  the  sale.  What 
was  the  rate  of  his  commission  ? 

17.  A  collector  remits  $1890  to  his  principal  after 
deducting  his  commission  of  10%.  What  was  the 
amount  collected? 

18.  A  farm  was  sold  for  $9384  at  a  commission  of 
1%.    Find  commission  and  proceeds  of  sale. 


46  ALGEBRAIC   ARITHMETIC 

19.  Kemitted  to  a  stockbroker  $  10,650,  to  be  invested 
in  stocks,  after  deducting  J%  brokerage.  What  amount 
of  stock  did  he  purchase? 

20.  A  broker  received  ^45,337  to  invest  in  bonds, 
after  deducting  a  commission  of  |^%.  What  amount 
did  he  invest,  and  what  was  his  commission? 

COMMERCIAL  DISCOUNT 

37.  Manufacturers  and  wholesale  dealers  avoid  the 
inconvenience  and  expense  of  issuing  price-lists  of  their 
goods  with  every  change  in  their  market  value  by  deter- 
mining upon  a  fixed  list  price  for  every  article  (largely  in 
excess  of  its  true  value),  from  which  they  give  their  cus- 
tomers certain  discounts,  determined  by  current  market 
prices. 

Goods  are  frequently  subject  to  two  or  more  discounts 
(the  last  generally  being  for  cash  payment)  ;  and  in  such 
cases  each  discount  is  reckoned  by  itself  on  the  sum 
remaining  after  subtracting  the  preceding  discounts. 

In  stating  commercial  discount,  the  sign  %  is  usually 
omitted. 

38.  Find  the  cost  of  a  bill  of  goods  amounting  to  $  800 
at  20  and  5  off,  and  5  off  for  cash. 

5)$  800      =  list  price  of  goods. 
160      =20%  discount. 
20)$  640 

32      =5%  discount. 

20)$  608 

30.40  =  5%  off  for  cash. 
$577.60  =  cost  of  the  goods. 


UNIVERSITY 

OF 

*-^      COMMERCIAL  DISCOUNT  47 

Or  as  follows : 

5  X  ?P  X  ^0 

EXAMPLES  15 

1.  Bought  goods  to  the  amount  of  $650  at  10  off, 
and  5  off  for  cash.     What  was  the  cost  ? 

2.  Find  the  cost  of  a  bill  of  goods  marked  at  $450 
at  40%  off,  and  5%  off  for  cash. 

3.  By  getting  a  discount  of  20,  and  10  off  for  cash,  I 
pay  $  1080  for  a  bill  of  goods.  What  was  the  list  price  ? 
What  single  discount  would  give  the  same  reduction  ? 

4.  For  what  must  I  sell  goods  which  were  sold  me 
for  $  830,  list  price,  at  30,  10,  and  5  off,  to  gain  20%  ? 

5.  Find  the  amount  of  a  bill  of  $  1560,  discounts 
being  40,  25,  and  5.    Find  the  single  equivalent  discount. 

6.  Sold  a  bill  of  goods  marked  at  $  250  for  30,  and 
5  off.  How  much  more  did  I  receive  than  if  I  had 
given  a  discount  of  35%  ? 

7.  Paid  $  655  for  a  bill  of  goods  after  a  discount  of 
16|%.     AVhat  was  the  invoice  price  ? 

8.  Find  the  cash  value  of  a  bill  of  cloth  amounting 
to  $425.50  at  a  discount  of  10%,  and  5%  off  for  cash. 
Find  the  equivalent  single  discount. 

9.  Find  the  cost  of  a  stove  listed  at  $  25,  discounts 
being  10  and  7^. 

10.   I  paid  $  1.50  for  a  book  after  a  discount  of  25%, 
and  16|  off.     What  was  its  marked  price  ? 


48  ALGEBRAIC   ARITHMETIC 

INSURANCE 

39.  Insurance  is  a  guaranty  to  pay  a  certain  sum  of 
money  in  case  of  loss  or  damage.  It  is  classed  as  insur- 
ance on.  property  and  insurance  on  life. 

That  on  property  is  called  fire  insurance,  if  against  loss 
by  fire;  marine  insurance,  if  against  loss  at  sea;  stock 
insurance,  if  against  the  loss  of  cattle,  horses,  etc. 

The  sum  paid  for  obtaining  the  insurance  is  called  the 
premium,  and  the  written  contract  is  called  the  policy. 

The  premium  is  a  certain  per  cent  of  the  sum  insured, 
and  is  paid  in  advance.  In  life  insurance  it  is  generally 
paid  annually. 

Fire-insurance  companies  rarely  insure  property  for 
more  than  two-thirds  of  its  value,  and  in  no  case  pay  for 
more  than  the  value  of  the  property  destroyed,  whatever 
may  be  the  face  of  the  policy. 

EXAMPLES  16 

1 .  What  is  the  premium  for  insuring  goods  for  $  14,500, 
at  1^%  ? 

2.  A  house  worth  $15,000  is  insured  for  |  of  its 
value,  at  |%.     What  is  the  premium  ? 

3.  A  ship  valued  at  $40,000  is  insured  for  |  of  its 
value,  at  1^%,  and  its  cargo,  valued  at  $36,000,  at  f  %. 
What  is  the  cost  of  insurance  ? 

4.  A  merchant  paid  $  1450  premium  for  the  insurance 
of  a  cargo  of  cotton,  the  rate  of  insurance  being  2|%. 
For  what  sum  was  the  cargo  insured  ? 

5.  If  it  cost  $  93.50  to  insure  a  store  for  ^t)f  its  value, 
at  1|%,  what  was  the  store  worth? 


INSURANCE  4y 

6.  A  merchant  pays  $  50  for  an  insurance  of  $  32,500 
on  a  shipment  of  goods  from  New  York  to  St.  Louis. 
What  is  the  rate  of  insurance  ? 

7.  A  house  valued  at  $  1200  had  been  insured  for  | 
of  its  value  for  3  years,  at  1%  per  annum.  During  the 
third  year  it  was  destroyed  by  fire.  What  was  the 
actual  loss  to  the  owner,  no  allowance  being  made  for 
interest  ? 

SuGGBSTiON.  The  difference  between  the  amount  of  the  insur- 
ance and  the  premium  for  the  three  years  is  what  he  gets  from  the 
insurance. 

8.  A  merchant  has  his  store  and  goods  insured  for 
$  5500  at  |-%  premium.  What  is  the  cost  to  him  ?  If 
the  store  and  goods  are  destroyed,  what  sum  does  the 
insurance  company  lose  ? 

9.  An  insurance  company  loses  $3528  by  the  wreck 
of  a  carload  of  flour  which  it  had  insured  for  $3600. 
What  was  the  rate  of  insurance  ? 

10.  A  merchant  insures  a  cargo  of  goods  for  $  81,800, 
which  sum  includes  the  value  of  the  goods  and  the  pre- 
mium at  2|%.  What  is  the  premium,  and  the  value 
of  the  goods  ? 

Suggestion.  The  premium  is  always  computed  on  the  amount 
of  insurance;  hence  in  this  case  the  base  is  $81,800. 

11.  A  merchant  ships  $  31,360  worth  of  wheat  from 
Chicago  to  Buffalo.  For  what  must  he  get  it  insured  at 
2%  so  as  to  cover  both  the  value  of  the  wheat  and  the 
premium  paid  for  its  insurance  ? 

Suggestion.  6  =  amount  of  insurance  =  $31,360  +  premium. 
Hence  $31,360  is  what  per  cent  of  6  ? 


50  ALGEBRAIC   ARITHMETIC 

12.  A  merchant  shipped  a  cargo  of  flour  worth  ^3597 
from  New  York  to  Liverpool.  For  what  must  he  insure 
it,  at  3J%,  to  cover  the  value  of  the  flour  and  the  pre- 
mium ? 

13.  I  insure  my  life  for  $8000,  paying  $19.80  per 
$  1000  per  year.  What  do  I  pay  the  company  if  I  live 
20  years  after  insurance  ? 

14.  The  annual  premium  on  a  life  insurance  at  2\% 
is  $  126.     What  is  the  amount  of  the  insurance  ? 

TAXES 

40.  A  tax  is  a  sum  of  money  assessed  upon  the  inhabi- 
tants of  a  town,  district,  county,  or  state,  or  upon  their 
property,  to  meet  some  public  expense,  such  as  the  sup- 
port of  the  schools,  or  of  the  government,  or  the  building 
of  public  works. 

A  tax  assessed,  without  regard  to  property,  upon  every 
male  citizen  within  certain  age  limits  (fixed  by  law)  is 
called  a  poll  tax,  or  capitation  tax.  A  person  so  assessed 
is  called  a  poll. 

A  property  tax  is  assessed  at  a  certain  per  cent  on  the 
estimated,  or  assessed,  value  of  taxable  property. 

Taxable  property  is  of  two  kinds :  (1)  Real  estate,  or 
fixed  property;  as  houses  and  lands;  (2)  Personal,  or 
movable  property;  as  furniture,  merchandise,  cattle, 
money,  etc. 

EXAMPLES  17 

1.  What  sum  must  be  assessed  to  raise  $83,600  net, 
after  deducting  the  cost  of  collection  at  5%  ? 

Remark.  The  cost  of  collection  is  5%  of  the  amount  collected. 
See  Art.  35,  second  paragraph,  and  (2)  of  the  Table. 


TAXES  51 

2.  In  a  certain  district,  a  school-house  is  to  be  built 
at  a  cost  of  $  18,500.  What  amount  must  be  assessed  to 
cover  this  and  the  collector's  fees  at  3%  ? 

3.  A  county  builds  a  bridge  for  $  4410.  The  property- 
is  valued  at  $1,000,000.  What  is  the  tax  per  $100, 
including  the  cost  of  collection  at  2%  ? 

4.  In  a  certain  town  a  tax  of  $5000  is  to  be  assessed. 
There  are  500  polls,  each  assessed  $  .75,  and  the  valua- 
tion of  the  taxable  property  is  $  370,000.  What  will  be 
the  rate  of  property  tax,  and  how  much  will  be  A's  tax, 
whose  property  is  valued  at  $7500,  and  who  pays  for 
2  polls? 

Suggestion.  Subtract  the  amount  to  be  raised  by  poll  tax  from 
the  whole  sum  to  be  assessed ;  and  find  the  per  cent  that  the 
remainder  is  of  the  value  of  the  taxable  property.  This  is  the  rate 
of  taxation. 

5.  A  tax  of  $  11,384,  besides  cost  of  collection  at  3J%, 
is  to  be  raised  in  a  certain  town.  There  are  760  polls 
assessed  at  $  1.25  each,  and  the  personal  property  is 
valued  at  $124,000,  and  the  real  estate  at  $350,000. 
Find  the  tax  rate,  and  find  a  person's  tax  whose  real 
estate  is  valued  at  $6750  and  personal  property  at 
$  2500,  and  who  pays  for  3  polls. 

6.  In  the  above  town,  how  much  is  B's  tax  on  $  15,000 
real  estate,  $  2750  personal  property,  and  2  polls  ? 

7.  What  is  C's  tax  on  $9786  and  1  poll  ? 

8»  How  much  taxes  will  a  person  pay  whose  property 
is  assessed  at  $  7500,  if  he  pays  f  %  town  tax,  ^%  state 
tax,  1 J  mills  on  a  dollar  school  tax  ? 


52  ALGEBKAIC   AUlTHMEriC 

9.  I  buy  a  lot  for  $  400  and  build  a  house  on  it  for 
$2000.  I  pay  an  insurance  on  the  house  of  |^%  on  J  of 
its  value,  and  a  tax  on  the  whole  of  14  mills  on  a  dollar, 
the  property  valuation  being  |  of  the  cost.  For  how 
much  must  I  rent  the  house  per  month  to  realize  10%  a 
year  on  my  money  ? 

10.  A  tax  of  $  56,000,  including  cost  of  collecting,  is  to 
be  raised  in  a  city  on  a  property  valuation  of  $  22,400,000. 
Assuming  that  the  uncollectible  tax  will  be  10%  of  the 
tax  assessed,  what  will  be  the  tax  rate  expressed  in  mills 
on  a  dollar  ? 

11.  In  the  above  city,  how  much  is  A's  tax  on  $  27,500  ? 

DUTIES 

41.  The  taxes  levied  on  imported  goods  are  called  cus- 
toms or  duties. 

Duties  are  of  two  kinds :  specific  and  ad  valorem. 

A  specific  duty  is  a  tax  on  goods  according  to  weight, 
number,  or  measure,  without  regard  to  value. 

An  ad  valorem  duty  is  a  percentage  of  the  cost  of  goods 
in  the  country  from  which  they  are  imported. 

Many  articles  are  subject  to  both  kinds  of  duty. 

Gross  weight  is  the  weight  of  goods  including  the  boxes 
or  other  packing  material. 

Net  weight  is  the  weight  after  deducting  the  weight  of 
the  packing  material. 

Specific  duties  are  calculated  on  the  net  weight  of 
goods.     All  custom-house  weights  are  long-ton  weights. 

The  following  list  is  taken  from  two  successive  tariffs 
of  the  United  States.  The  new  superseded  the  old  July 
24,1897: 


DUTIES 


53 


ABTI0LK8 

Old  Kate  of  Duty 

N::\v  Rate  of  Duty 

Alcoholic  perfumery 

$2  per  gal.  and  50% 

$.60  per  lb.  and  45% 

Earthen  and  crock- 

ery ware  .... 

30% 

55% 

Glass,  cut,  engraved, 

or  painted    .    .    . 

35% 

60% 

Tin  plate     .    .    .    . 

U  ct.  per  lb. 

1|  ct.  per  lb. 

Machinery  .... 

35% 

45% 

Cigars 

$4  per  lb.  and  25% 

$4.50  per  lb.  and  25% 

Horses     valued     at 

$150  or  less     .    . 

20% 

$30  per  head 

Wheat 

20% 

25  ct.  per  bu. 

Cotton  clothing, 

ready-made  .    .    . 

40% 

50% 

Cotton  hosiery,  val- 

ued at   not   more 

than  $1  per  doz. 

pairs 

50% 

$.50 per  doz.  and  15% 

Shirts  and  drawers 

valued  at  not  more 

than  $1.50  per  doz. 

60% 

$  .60  per  doz.  and  15% 

Collars  and  cuffs  of 

linen 

30ct.  perdoz.  and3% 

40  ct.  per  doz.  and  20% 

Laces  and  embroid- 

eries of  linen    .    . 

50% 

60% 

Silk  velvets     .    .    . 

$1.50  per  lb. 

$1.50  per  lb.  and  15% 

Lead  pencils    .    .    . 

50% 

45  ct.  per  gross  and  25% 

EXAMPLES  18  (Oral) 

1.  Which  is  the  higher  duty  on  horses  valued  at 
$  150  ?     Less  than  $  150  ? 

2.  For  what  value  of  wheat  are  the  two  duties  on  that 
article  equal  ?  Which  is  the  higher,  and  by  how  much, 
when  wheat  is  worth  $  .60  ? 


54  ALGEBRAIC   ARITHMETIC 

3.  What  is  the  difference  between  the  new  and  old 
duties  on  $1000  worth  of  cut  glass?  On  $25,000 
worth  of  machinery? 

4.  What  is  the  cost  per  gross  of  lead  pencils  on  which 
the  two  rates  of  duty  are  equal  ?  Which  is  the  greater 
for  pencils  worth  more  than  that?  For  pencils  worth 
less? 

5.  What  was  the  old  duty  on  $  5000  worth  of  ready- 
made  clothing  ?    What  is  the  new  ? 

6.  What  is  the  new  duty  on  100  lb.  of  perfumery 
worth  $  2  per  pound  ? 

7.  What  is  the  new  duty  on  a  dozen  collars  valued  at 
$  1.20  per  doz.  ? 

8.  What  is  the  difference  between  the  old  duty  and 
the  new  on  a  ton  of  tin  plate  ? 

EXAMPLES  19 

Find  the  (new)  duty : 

1.  On  1000  boxes  of  cigars,  each  containing  100  cigars, 
invoiced  at  $  7.25  per  box.     Net  weight  12  lb.  per  1000. 

2.  On  12  gross  lead  pencils  at  $1.00  per  gross. 

20  gross  lead  pencils  at  $  2.25  per  gross. 
5  gross  lead  pencils  at  $  5.00  per  gross. 

3.  On  machinery  invoiced  at  $  26,500. 

4.  On  150  yd.  silk  velvet  at  $  1.75  per  yd.  Net  weight 
75  1b. 

5.  On  15  doz.  shirts  at  $  1.50  per  doz. 

20  doz.  linen  collars  at  $1.10  per  doz. 
50  yd,  linen  lace  at  5  ^  per  yd. 
7  doz.  cotton  hose  at  $  .90  per  doz. 


MISCELLANEOUS  EXAMPLES  ,65 

EXAMPLES  20  (Miscellaneous) 

1.  A  man  had  $5420  in  bank.  He  drew  out  15%  of 
it,  then  20%  of  the  remainder,  and  afterwards  deposited 
121%  of  what  he  had  drawn.  How  much  had  he  then 
in  bank  ? 

2.  If  a  man  owning  45%  of  a  steamboat  sells  16|% 
of  his  share  for  $  5860,  what  is  the  value  of  the  whole 
boat? 

3.  A  man  sold  two  houses  at  $2500  each;  on  one  he 
gained  20%,  on  the  other  he  lost  20%.  What  was  his 
loss  on  the  two  sales  ? 

4.  A  man  bought  a  piece  of  property  which  after- 
wards increased  in  value  each  year  at  the  rate  of  25% 
on  the  value  of  the  previous  year,  for  4  years ;  and  was 
then  worth  $16,000.     What  did  it  cost  ? 

5.  After  deducting  6^%  commission  and  $132  for 
storage,  my  agent  sends  me  $  23,654.25  as  the  net  pro- 
ceeds of  a  consignment  of  pork  and  flour.  What  was  the 
amount  of  the  sale  ? 

6.  After  taking  out  15%  of  the  grain  in  a  bin,  there 
remained  40  bu.  3^  pk.  How  many  bushels  were  there 
at  first  ? 

7.  The  profits  of  a  farm  in  2  years  were  $3485,  and 
the  profits  of  the  second  year  were  5  %  greater  than  those 
of  the  first  year.    What  were  the  profits  of  each  year  ? 

8.  If  |-  of  a  farm  is  sold  for  what  f  of  it  cost,  what  is 
the  gain  per  cent  ? 

9.  What  is  the  cost  of  goods  sold  for  $47,649,  at  a 
profit  of  16}%  ? 


56,  ALGEBRAIC   ARITHMETIC 

10.  A  broker  receives  $  7125  to  invest  in  cotton,  after 
deducting  his  commission  of  2|^%.  How  many  pounds 
of  cotton  can  he  buy  at  11^  cents  a  pound  ? 

11.  Sold  a  farm  for  $14,700,  and  lost  12%.  What 
per  cent  should  I  have  gained  by  selling  it  for  $  21,000  ? 

12.  I  buy  a  house  for  $6500  and  spend  $500  for 
repairs.  I  rent  it  for  $  77.50  a  month,  out  of  which  I 
pay  a  yearly  insurance  of  |%  on  -f-  of  its  whole  cost, 
including  repairs,  and  a  yearly  tax  of  1%  on  J  of  the 
same.  What  per  cent  of  income  a  year  do  I  realize  on 
the  whole  cost  ? 

13.  For  what  sum  must  a  policy  be  made  out  to  cover 
the  insurance  on  property  worth  $2100,  at  |^%  ? 

14.  I  bought  a  lot  of  coffee  at  12^  per  pound.  Allow- 
ing that  the  coffee  will  fall  5%  short  in  weighing  it  out, 
and  that  10%  of  the  sales  will  be  in  bad  debts,  for  how 
much  per  pound  must  I  sell  it  to  make  a  clear  gain  of 
14%  on  the  cost? 

15.  An  agent  sells  for  Johnson  &  Co.  3500  lb.  of  butter 
at  20^  per  pound,  and  2580  lb.  of  cheese  at  9^  per  pound, 
at  a  commission  of  5%.  He  invests  the  balance  in  dry 
goods,  after  deducting  his  commission  of  2J%  for  pur- 
chasing. How  many  dollars'  worth  of  goods  do  Johnson 
&  Co.  receive?  What  is  the  entire  commission  of  the 
agent  ? 


CHAPTER  IV 

APPLICATIONS   OF   PERCENTAGE   INVOLVING  TIME 

42.  The  money  paid  for  the  use  of  money  is  called 
interest.     It  is  always  a  percentage  of  the  sum  loaned. 

The  sum  loaned  is  called  the  principal. 

The  rate  per  cent  of  the  principal  paid  for  its  use  for 
a  certain  time  is  called  the  rate  of  interest.  It  is  under- 
stood to  be  for  a  year  unless  otherwise  specified. 

The  sum  of  the  principal  and  the  interest  is  called  the 
amount.  It  is  the  sum  that  the  borrower  must  pay  back 
to  cancel  his  debt. 

In  computing  interest  for  a  fraction  of  a  year,  it  is 
customary  to  reckon  each  month  as  ^^  of  a  year,  and  a 
day  as  -^^  of  a  month. 

SIMPLE   INTEREST 

43.  Ex.  1.  What  is  the  interest  on  $100  for  2^  yr. 
at8%? 

The  interest  for  1  yr.  is  S%  of  $100,  or  $8,  and  for 
21  yr.  is  21  X  $  8,  or  $  20.     Or, 

Since  the  interest  is  8%  of  the  principal  for  1  yr.,  for 
21  yr.  it  will  be  2i-  x  8%  or  20%  of  the  principal;  and 
20%  of  $100  =  $20. 

Ex.  2.  Find  the  interest  and  amount  of  $200  for  3  yr. 
at  5%. 

67 


58  ALGEBRAIC   ARlTIiMETIC 

The   interest  is   15%   of  the  principal,  or  $30;   the 
amount  is  $  200  +  $  30  =  $  230. 


EXAMPLES  21  (Oral) 

rind  the  interest  and  amount  of : 

3.  $  100,  at  6%,  for  1  yr. ;  2^  yr. ;  3  yr.  4  mo. 

4.  f  500,  at  5%,  for  6  mo. ;  2  yr. ;  2^  yr. 

5.  $50,  at  12%,  for  1  mo.;  1^  yr. 

6.  $1000,  at  3%,  for  1  yr. ;  2  yr.  4  mo. 

7.  $40,  at  6%,  for  2  mo.;  6  mo.;  10  mo. 

8.  $  5  at  10%,  for  2  yr. ;  3|  yr. 

9.  $  10  at  6%,  for  1  mo. ;  9  mo. 

10.  $  300  for  6  mo.,  at  6%  ;  at  8%. 

11.  $60  for  8  mo.,  at  6%;  at  12%. 

12.  $  200  for  3  mo.,  at  1%  a  month. 

13.  $  250  for  2|  yr.,  at  4%  ;  at  10%. 

14.  $  6  for  7  mo.,  at  1%  a  month. 

44.  To  compute  interest  at  6  per  cent.  Keckoning  a 
month  SiS  ^  oi  Si  year  and  a  day  as  ^^  of  a  month,  the 
interest,  at  6%, 

for  1  yr.  =  .06  of  the  principal ; 
for  2  mo.  =  .01  of  the  principal ; 
for  1  mo.  =  .005  of  the  principal; 
for  6  da.  =  .001  of  the  principal ; 
.  for  1  da.  =  .000^  of  the  principal. 


SIMPLE  INTEREST  59 

Hence  to  find  the  decimal  fraction  of  the  principal  that 
the  interest,  at  6%,  for  any  given  time  is,  take  6  times  the 
number  of  year's  and  ^  the  number  of  months  as  hundredths^ 
and  \  the  number  of  days  as  thousandths. 

Ex.  1.   Find  the  interest  of  ^375.50  for  3  yr.  5  mo. 
21  da.,  at  6%. 
The  interest  for 

3  yr.  =    3  X  .06      =  .18      of  the  principal ; 

5  mo.  =    5  X  .005    =  .025    of  the  principal ; 

21  da.  =  21  X  .000^  =  .0035  of  the  principal; 

.2085  of  the  principal. 

Or,  following  the  rule  exactly, 

^375.5 
(3  X  .06)  =  .18  -2085 

(I  X  .01)  =  .025  7510 

(Vx. 001)  =.0035  300 

.2085  19 

$78.29   interest. 

Note.  In  forming  the  multiplier,  the  operations  indicated  in 
parentheses  should  be  performed  mentally,  only  the  results  being 
set  down. 

EXAMPLES  22 

Find  the  interest  and  amount,  at  6%,  of : 

2.  $  760,  for  1  yr.  9  mo.  27  da. 

3.  $  179.50,  for  1  yr.  1  mo.  8  da. 

4.  $  325,  for  2  yr.  11  mo.  6  da. 

6.   $  758.75,  for  3  yr.  2  mo.  16  da. 


60  ALGEBRAIC   ARITHMETIC 

6.  $  1024.25,  for  2  yr.  3  mo.  22  da. 

7.  $  584.50,  for  1  yr.  2  mo.  14  da. 

8.  $  725.84,  for  1  yr.  3  mo.  11  da. 

9.  $  387.95,  for  3  yr.  7  mo.  24  da. 
10.  $  42.20,  for  24  da. 

45.  To  find  the  years,  months,  and  days  between  two 
dates,  add  mentally  to  the  earlier  date  first  the  years, 
then  the  months,  then  the  days  necessary  to  obtain  the 
later  date,  in  each  case  recording  only  the  result. 

Ex.   Find  the  time  from  Jan.  26, 1895,  to  June  8, 1897. 

From  Jan.  26, 1895,  to  Jan.  26, 1897,  2  yr. ;  to  May  26, 
4  mo. ;  to  June  8,  12  da.  (4  in  May  and  8  in  June). 
Time:   2  yr.  4  mo.  12  da. 

Note.  Observe  that  (1)  the  last  day  is  counted,  the  first  is  not ; 
(2)  where  in  counting  the  days  we  pass  from  one  month  to  the 
next,  the  whole  number  of  days  in  the  former  is  taken  as  thirty 
for  any  month  of  the  year. 

46.  For  rates  other  than  6  per  cent,  the  multiplier  is  most 
readily  found  by  the  six  per  cent  method,  as  follows : 

First  find  the  multiplier  for  the  given  time,  at  6% ; 
then 

for  3%  take  ^  of  it ;  for  7%  add  ^ ; 

for  4%  subtract  J  of  it ;  for  8%  add  |; 

for  5%  subtract  ^ ;  for  9%  add  ^,  etc. 

For  rates  higher  than  10%,  it  is  easier  to  form  a  12% 
multiplier  with  the  months  as  hundredths,  and  ^  the 
days  as  thousandths. 


SIMPLE  INTEREST  61 

EXAMPLES  23 

Find  the  interest  on : 

1.  $  721.56,  for  1  yr.  4  mo.  10  da.,  at  6%. 

2.  $  54.75,  for  3  yr.  24  da.,  at  5%. 

3.  $  1000,  for  11  mo.  18  da.,  at  7%. 

4.  $3046,  for  7  mo.  26  da.,  at  8%. 

5.  $  1821.50,  from  April  1  to  Nov.  12,  at  6%. 

6.  $  700,  from  Jan.  15  to  Aug.  1,  at  10%. 

7.  $316.84,  from  Oct.  20,  1895,  to  March  10,  1897, 
at  7%. 

8.  $127.36,  from  Dec.  12,  1893,  to  July  3,  1895, 

at  4^%. 

Find  the  amount  of : 

9.  $  3146,  for  2  yr.  3  mo.  10  da.,  at  7%. 

10.  $  1008.80,  for  10  mo.  16  da.,  at  6^%. 

11.  $  2000,  for  15  da.,  at  12^%. 

12.  $  137.60,  for  127  da.,  at  10%. 
Note.    Count  30  da.  to  a  month. 

13.  $  1671.64,  from  June  1,  1894,  to  April  1,  1896, 
at  7%. 

14.  $  250,  from  June  5, 1896,  to  Feb.  14, 1897,  at  8%. 

15 .  $  340.50,  from  May  25, 1895,  to  Sept.  9, 1897,  at  9  % . 

16.  $25,  for  93  da.,  at  12%. 

17.  $  145.20,  for  1  yr.  11  mo.  29  da.,  at  7%. 


62  ALGEBRAIC   ARITHMETIC 

18.  $  450,  for  3  yr.  2  mo.  21  da.,  at  8%. 

19.  A  man  engaged  in  business  was  making  12^% 
annually  on  his  capital  of  $  16,840.  He  quit  his  business, 
and  loaned  his  money  at  T-|-%.  What  did  he  lose  in 
2  yr.  3  mo.  18  da.  by  the  change  ? 

20.  A  speculator  borrowed  $9675,  at  6%,  April  15, 
1894,  with  which  he  purchased  flour  at  $  6.25  per  bbl. 
May  10,  1895,  he  sold  the  flour  at  $  7|  per  bbl.,  cash. 
What  did  he  gain  by  the  transaction  ? 

21.  A  man  borrows  $1000  at  10%  interest,  and  with 
it  buys  a  note  for  $  1100,  maturing  in  5  mo.,  but  which 
not  being  paid  when  due,  runs  1  yr.  6  mo.  beyond  matu- 
rity, drawing  interest  at  6%  after  maturity.  How  much 
does  he  gain  ? 

47.  Accurate  Interest.  The  common  method  of  com- 
puting interest  is  accurate  for  whole  years;  but  is  not 
accurate  for  months,  since  no  month  is  exactly  -^^  of  a 
year ;  nor  for  days,  since  a  day  is  reckoned  as  -g^-^  of  a 
year. 

To  compute  accurate  interest,  find  the  interest  for 
years  by  the  common  method ;  and  for  any  fraction  of  a 
year,  take  as  many  365ths  of  a  year's  interest  as  there 
are  days. 

Note.  The  number  of  days  in  each  month  can  be  remembered 
from  the  following : 

"  Thirty  days  hath  September, 
April,  June,  and  November : 
All  the  rest  have  thirty-one, 
Except  the  second  month  alone, 
Which  has  but  twenty-eight,  in  fine, 
Till  leap-year  gives  it  twenty-nine." 


SIMPLE  INTEREST  63 

Ex.  1.  Find  the  accurate  interest  on  $535  from  July 
25  to  Oct.  3,  at  6%. 

Time :    July,     6  da. 
Aug.,  31 

Sept.,  30  $535x6x70  ^  ^  ^^^  ^^^^ 

Oct.,      3  10  X  365 

70  da. 

EXAMPLES  24 

2.  Find  the  interest  in  Ex.  1  by  the  common  method. 

3.  Find  the  exact  interest  on  3  United  States  bonds,  of 
$  1000  each,  at  6%,  from  May  1  to  Oct.  15. 

4.  What  is  the  exact  interest  on  a  $  500  United  States 
bond,  at  5%,  from  Nov.  1  to  April  10? 

Find  the  exact  interest  on : 

5.  $375,  from  June  12,  1896,  to  Dec.  14,  1897,  at  7%. 

6.  $  5760,  from  Nov.  8, 1896,  to  March  1, 1897,  at  6%. 

7.  1 12,085,  from  Sept.  4,  to  Dec.  17,  at  5%. 

8.  $  1250,  from  April  1,  to  Dec.  7,  at  6%. 

9.  What  is  the  difference  between  the  exact  interest 
for  90  da.  on  $1,000,000  of  6%  bonds  and  the  interest 
reckoned  on  the  basis  of  360  da.  to  the  year  ? 

PKOBLEMS  IN  INTEREST 

48.  Interest  Formulas.  Since  the  interest  and  the 
amount  in  any  problem  in  interest  are  percentages  of  the 
principal,  the  relation  that  these  quantities  bear  to  one 


64  ALGEBRAIC   ARITHMETIC 

another  can  easily  be  expressed  by  the  formulas  of 
Art.  32. 

Thus,  applying  (1)  Art.  32  to  Ex.  2,  Art.  43,  we  see 
that: 

To  find  the  interest y  6  =  $  200,  r  =  .15. 

Hence  p  (the  interest)  =  $  200  x  0.15  =  $  30. 

To  find  the  amount,  &  =  $  200,  r  =  1.15. 

Hence  p  (the  amount)  =  $  200  x  1.15  =  f  230. 

It  will  be  seen  that,  in  finding  the  interest,  r  is  the 
product  of  two  factors ;  namely,  the  rate  of  interest  and 
the  time.  We  shall  obtain  much  more  useful  formulas 
by  denoting  each  of  these  factors  by  a  letter,  and  by 
always  using  the  same  letters  to  denote  the  same  ele- 
ments. For  this  purpose  we  shall  use  the  initials  of  the 
names  of  the  elements. 

Thus,  p  =  the  principal  (base), 

r  =  the  rate  of  interest  (per  annum,  unless  other- 
wise specified), 

t  =  the  time  (expressed  in  the  same  denomina- 
tion as  that  for  which  the  rate  is  given), 

i  =  the  interest, 

a  =  the  amount  =:p-\-i. 

Note.  It  should  be  observed  thatp  is  not  here  a  percentage  of 
some  number,  as  in  the  preceding  formulas.  It  is  the  base,  of 
which  i  and  a  are  percentages. 

49.  The  formula  expressing  the  rule  by  which  we 
have  computed  the  interest  in  all  preceding  examples  is 

''  =  prt-,  (1) 

in  which  the  product  rt  is  the  multiplier  (Art.  46). 


PROBLEMS  65 

This  equation  involves  four  quantities,  any  one  of 
which  can  be  found  if  the  other  three  are  given.  For 
we  have  only  to  solve  the  equation  for  the  unknown 
quantity ;  then  substitute  the  given  values  of  the  other 
quantities,  and  perform  the  indicated  operations. 

Ex.  1.  What  sum  of  money  will  gain  $84  interest  in 
2  yr.,  at  7%  ? 

1=  $84,  r  =  .07,  t  =  2j  p  =  ? 

Solve  (1)  for  p  by  dividing  its  sides  by  rtf  and  inter- 
change the  sides ;  then 

p-if  (2) 

Hence  ^  =  ^^^=$600.    Ans, 

.14 

NoTB.  The  value  of  p  is  the  answer  to  the  question :  84  is  14  % 
of  what  number  ?     (Case  III.  Art.  32.) 

Ex.  2.   At  what  rate  will  $300  gain  $  60  in  4  years  ? 
p  =  $  300,  t  =  $  60,  ^  =  4,  r  =  ? 
Divide  the  members  of  (1)  by  pt ;  then 

r  =  l.  (3) 

Analysis.  The  interest  on  $  300  for  4  yr.  is  the  same  as  the 
interest  on  |1200  for  one  year,  the  rate  remaining  the  same. 
Hence  the  rate  is  yf^  ^ ,  or  5  %. 

Note.  The  t  in  all  the  interest  formulas  is  really  an  abstract 
multiplier;  its  value  is  the  number  of  years.  The  product  of 
dollars  and  years  is,  of  course,  impossible. 


66  ALGEBRAIC   ARITHMETIC 

Second  solution  and  analysis : 
15 

$300  x^     "^ 

The  interest  for  1  yr.  is  \  the  interest  for  4  yr.,  or 
$  15.    Hence  the  rate  is  -^^-^^  or  5%. 

Ex.  3.   In  what  time  will  %  500  gain  $  60,  at  4%  ? 

p  =  $500,  i  =  $60,  r  =  .04,  t=? 
Solving  (1)  for  ty  we  have 

Hence  t  =  ^,,f  ^^^^^  =  ^  =  3.  3  yr.  Ans. 

$500x0.04     20  ^ 

Analysis.  For  1  yr.  the  interest  is  %  500  x  0.04,  or  $20.  Hence 
to  gain  %  60  it  will  take  as  many  years  as  $  20  is  contained  times 
in  $60,  or  3  yr. 

60.  From  the  definition  of  amount  we  have  the  for- 
mula 

a=p  +  i.  (1) 

Hence,  replacing  i  by  its  value  prt, 

a=p-\-prf,  (2) 

or  a=p{l  +  rf),  (3) 

Equation  (3)  expresses  the  fact  that  the  amount  may 
be  obtained  directly  from  the  principal  by  adding  1  to 
the  multiplier  by  which  the  interest  is  obtained. 

Ex.  1.   Find  the  amount  of  $  250  for  S^  yr.,  at  6%. 
The  interest  is  15%  of  the  principal ;  hence  the  amount 


PROBLEMS  67 

is  115%  of  it.     The  multiplier  for  the  interest  is  .15; 
for  the  amount  it  is  1  +  .15,  or  1.15. 

a  =  1 250  X  1.15  =  $  287.5.  (Case  I.) 

Ex.  2.    What  principal  will  amount  to  $267.90  in 

2yr.,at7%? 

a  =  $267.90,  r  =  .07,  t  =  2,  p=? 
Solve  (3)  for  p  by  dividing  its  sides  by  (1  +  rt) ;  then 

Hence       p  =  f|^  =  »|^  =  $  235.  (Case  III.) 

Subtract  p  from  both  sides  of  (2) ;  then 
a—p=:  prtf 
or  prt=a—p.  (6) 

Divide  both  sides  of  (5)  hy  pt-,  then 

.  =  ^.  (6) 

But  a—p  =  i. 

Hence,  substituting,    r  =  — : 

pt 

which  is  the  same  as  (3)  Art.  49. 

Divide  the  members  of  (5)  by  pr ;  then 


pp 


(7) 


68  ALGEBRAIC   ARITHMETIC 

Replacing  a—phy  t, 

<  =  -; 

which  is  the  same  as  (4)  Art.  49. 

51.   Interest  Formulas.     From  the  last  two  articles  we 

obtain  the  following  set  of  interest  formulas.    References 
are  given  to  the  corresponding  percentage  formulas. 

Given  the  Principal,  Rate,  and  Time ;  to  find  the  Interest. 

/  =  pri.  (Case  I.)    (1) 

Given  the  Principal,  Rate,  and  Time;  to  find  the  Amount. 

Given  the  Interest  (or  Amount),  Principal,  and  Time;  to 
find  the  Rate. 

/^  .  ''*_        \    (Case  II.)    (3) 

Given  the  Interest,  Rate,  and  Time ;  to  find  the  Principal. 
p=^'  (Casein.)    (4) 

Given  the  Amount,  Rate,  and  Time ;  to  find  the  Principal. 

P  =  Yf7i  (Casein.)    (5) 

Given  the  Interest  (or  Amount),  Principal,  and  Rate;  to 
find  the  Time. 

t  =  J-,  i  =  ^^zJL.  (None.)    (6) 

pr  pp 


PROBLEMS  69 

EXAMPLES  25  (Oral) 
Find : 

1.  Sum  that  will  gain  $  20  in  2  yr.,  at  5%. 

2.  Sum  that  will  amount  to  $  228  in  2  yr.,  at  7%. 

3.  Rate  at  which  $  400  will  gain  $  84  in  3  yr. 

4.  Time  in  which  $  200  will  gain  ^  56  at  7%. 

5.  Rate  at  which  $  120  will  gain  $  60  in  10  yr. 

6.  Sum  that  will  amount  to  $  350  in  15  yr.,  at  5%. 

7.  Time  in  which  $  1000  will  gain  $  250  at  5%. 

8.  Sum  that  will  gain  $  900  in  3  yr.,  at  3%. 

9.  Rate  at  which  $  5  will  gain  ^  1  in  3  yr. 

10.  Sum  that  will  amount  to  $  260  in 3  yr.  9  mo.,  at  8%. 

11.  Time  in  which  $  100  will  gain  $  15,  at  6%. 

12.  Rate  at  which  $  50  will  gain  ^  1.50  in  6  mo. 

13.  Sum  that  will  gain  $  40  in  6  mo.,  at  8%. 

14.  Rate  at  which  $  200  will  gain  $  25  in  2^  yr. 

15.  Time  in  which  $  75  will  gain  $  5,  at  4%. 

16.  Time  in  which  ^150  will  gain  $  21,  at  8%. 

17.  Rate  at  which  $  400  will  amount  to  ^  460  in  2 J  yr. 

18.  Time  it  will  take  $  700  to  amount  to  $  749,  at  7%. 

19.  Rate  at  which  any  sum  will  double  itself  in  10  yr. ; 
in  8  yr.  4  mo. ;  in  16  yr. ;  in  20  yr. 

20.  Time  it  will  take  money  to  double  itself,  at  5%;  at 
6%;  at  8%. 


70  ALGEBRAIC   ARITHMETIC 

EXAMPLES  26 
Find: 

1.  The  principal  tliat  will  gain  ^  213  in  5  yr.  10  mo. 
20  da.,  at  7%. 

2.  Sum  that  will  amount  to  $1028  in  4  mo.  24  da., 
at  7%. 

3.  Time  in  which  $  1301.64  will  amount  to  f  1522.92, 
at  5%. 

4.  Kate  at  which  $  1350  will  amount  to  $  1539  in  2  yr. 
4  mo. 

5.  Sum  that  will  amount  to  $  761.44  in  3  yr.  4  mo. 
24  da.,  at  5%. 

6.  Rate  at  which  $  1500  will  gain  $  252  in  2  yr.  4  mo. 
24  da. 

Note.  (  =  2f  yr.  When  the  months  and  days  are  not  expres- 
sible as  a  fraction  with  small  terms,  it  is  simpler  to  form  a  1%  mul- 
tiplier for  the  given  time,  and  solve  by  the  second  set  of  formulas  (3). 
Solve  Ex.  6  by  both  methods. 

7.  Time  in  which  $  175.12  will  gain  $  6.43,  at  6%. 

8.  Kate  at  which  $2085  will  gain  $68.11  in  5  mo. 
18  da. 

9.  Sum  that  will  gain  $  173.97  in  4  yr.  4  mo.,  at  69^. 

10.  Sum  that  will  amount  to  $  1596  in  2  yr.  6  mo., 
at  5J%. 

PRESENT  WORTH  AND  TRUE  DISCOUNT 

52.  The  present  worth  of  any  debt  is  the  sum  or  prin- 
cipal which  at  the  current  rate  of  interest  will  amount  to 
that  debt  when  it  becomes  due. 


PRESENT  WORTH  AND   TRUE  DISCOUNT  71 

The  difference  between  the  amount  of  the  debt  and  its 
present  worth  is  called  the  true  discount. 

Problems  in  present  worth  and  true  discount  are  solved 
by  formula  (5) ;  in  which  a  is  the  amount  of  the  debt, 
andp  its  present  worth  (base). 

Ex.  1.  A  merchant  buys  a  bill  of  goods  for  $  700  on  3 
months'  time.  What  is  the  present  worth  of  the  debt, 
money  being  worth  6%  ? 


P  = 


iI55L_  =  iI22=$  689.66.  Ans. 


l-^rt     1+  0.015      1.015 


EXAMPLES   27 

2.  What  is  the  present  worth  and  discount  of  a  debt 
of  $  1000  due  in  1  yr.  6  mo.,  the  current  rate  of  interest 
being  6%  ? 

3.  A  merchant  buys  goods  for  $4200  on  4  mo.  credit, 
but  is  offered  a  discount  of  3%  for  cash.  If  money  is 
worth  ^%  a  month,  what  is  the  difference  ? 

4.  Bought  a  house  and  lot  for  $19,500  cash,  and  sold 
.them  for  $22,000,  payable  ^  in  cash  and  the  remainder 
in  1  yr.  6  mo.  How  much  did  I  gain,  computing  dis- 
count at  6%  ? 

5.  A  merchant  holds  two  notes,  one  for  $356.25,  due 
Dec.  1,  1897,  and  the  other  for  $497.50,  due  Feb.  1, 1898. 
What  would  be  due  him  in  cash  on  both  notes  Sept.  15, 


1897,  discounting  at  6%  ? 

6.  Which  is  the  more  profitable,  to  buy  coal  at  $8.75 
per  ton  on  6  mo.  credit,  or  at  $8.60  on  2  mo.  credit, 
money  being  worth  7%  ? 


72  ALGEBRAIC    ARITHMETIC 

7.  What  sum  must  I  put  at  interest  at  8%  to  liquidate 
a  debt  of  $  2500  due  3  yr.  hence  ? 

8.  Bought  a  house  for  $  7500,  payable  in  4  mo.,  and 
sold  it  for  $7500  cash.  If  money  is  worth  6%,  what 
did  I  gain? 

9.  Find  the  difference  between  the  interest  and  true 
discount  of  $270  for  9  mo.,  at  8%. 

BANK  DISCOUNT 
53.  Promissory  Notes. 

$  150.00.  Cambridge,  Mass.,  July  28,  1897. 

Ninety  days  after  date,  I  promise  to  pay  Charles  Bond 
One  Hundred  and  Fifty  Dollars,  value  received. 

John  Brainard. 

$2000.00.  Berkeley,  Cal.,  May  15,  1896. 

Sixty  days  after  date,  I  promise  to  pay  to  the  order 

of  Frank  Barnes  Two  Thousand  Dollars,  value  received, 

with  interest  at  6%. 

W.  B.  Slack. 

The  above  are  examples  of  promissory  notes  ;  so  called 
because  they  contain  a  promise  to  pay  a  certain  sum,  at 
a  specified  time,  for  value  received. 

The  person  who  signs  a  note  is  called  the  maker; 
the  person  to  whom  or  to  whose  order  it  is  to  be  paid 
is  called  the  payee.  The  sum  named  in  a  note  is  called 
its  face. 

In  most  states  a  note  is  not  legally  due  till  three  days 
after  the  expiration  of  the  time  specified  in  the  note. 


BANK  DISCOUNT  73 

These  are  called  days  of  grace.  They  are  counted  in 
by  bankers  in  discounting  notes.  A  note  is  said  to 
mature  on  the  last  day  of,  grace. 

To  find  the  date  of  maturity  when  the  time  is  ex- 
pressed in  days,  count  forward  from  the  date  of  the  note, 
the  specified  number  of  days  plus  the  days  of  grace, 
reckoning  the  actual  number  of  days  in  the  months 
passed  over.  When  the  time  is  expressed  in  months, 
calendar  months  are  always  to  be  understood. 

A  note,  like  the  second  given  above,  made  payable  to 
the  order  of  the  payee,  or  one  made  payable  to  hearer, 
is  a  negotiable  note ;  that  is,  it  can  be  bought  and  sold. 

54.  Bank  Discount.  If  the  holder  of  a  negotiable  note 
sells  it  to  a  bank,  he  will  receive  the  amount  of  the  note 
at  maturity  minus  a  percentage  of  that  sum,  called  the 
bank  discount,  which  is  computed  at  a  certain  per  cent 
per  month  or  per  annum. 

The  sum  received  for  the  note  is  called  the  proceeds 
or  avails  of  the  note. 

Note.  Banks  reckon  12  mo.  of  30  da.  each,  or  360  da.  to  a 
year;  and  count  the  actual  number  of  days  in  a  given  time. 

Ex.  1.  Find  the  discount  and  proceeds  of  the  first  note 
in  the  last  article,  if  discounted  at  a  bank  Aug.  31,  at  1  % 
per  month. 

The  term  of  discount  =  93  da.  —  (3  da.  in  July  4-  31 
da.  in  Aug.)  =  59  da. 

Discount  =  interest  on  $  150  f or.59  da.  at  1  %  per  mo. 

=  $  150  x0.019|  =  f  2.95. 
Proceeds  =  $  150  -  $  2.95  -=  $  147.05. 


74  ALGEBRAIC   ARITHMETIC 

Ex.  2.  Find  the  discount  and  proceeds  of  the  second 
note,  discounted  on  the  day  it  was  made,  at  6%. 

Interest  on  $  2000  for  2  mo.  3  da.  =  f  2000 x. 0105  =  ^21. 
Amount  of  note  at  maturity  =  $  2021. 
Discount  =  interest  on  $  2021  for  2  mo.  3  da.  =  $  21.22. 
Proceeds  =  $  2021  -  $  21.22  =  ^  1999.78. 

Note.  The  present  worth  of  the  note  at  6%  discount  is,  of 
course,  its  face  ;  and  the  true  discount  on  the  amount  at  maturity 
is  $21.  The  excess  of  the  bank  discount  above  the  true  discount 
is  equal  to  the  interest  on  the  true  discount  for  the  given  time 
($21x.0105  =  $.22). 

EXAMPLES  28 

Find  the  date  of  maturity,  the  term  of  discount,  and 
the  proceeds  of  the  following : 

3.  $  OST^^j^.  Boston,  July  27,  1897. 

Three  months  after  date,  I  promise  to  pay  to  the  order 
of  M.  Levering  Nine  Hundred  Fifty-seven  and  rf^  Dol- 
lars, value  received. 

Discounted  Aug.  10,  at  8%.  T.  J.  Jennings. 

4.  $916^^.  Glendalb,  Cal.,  Feb.  5,  1896. 

Two  months  after  date,  we  jointly  and  severally 
promise  to  pay  C.  R.  Crowley,  or  order,  Nine  Hundred 
Sixteen  and  y^^  Dollars,  value  received,  with  interest 
at  8%. 

Discounted  Feb.  21,  at  10%.  James  Little. 

T.  B.  Long. 


ANNUAL   INTEREST  76 

5.  $700.00.  New  York,  April  10, 1897. 
Four  months  after  date,  I  promise  to  pay  to  the  order 

of  Edward  Brill  Seven  Hundred  Dollars,  value  received. 
Discounted  June  10,  at  8%.  ^    ^    Gorden. 

Write  the  following  in  the  form  of  promissory  notes, 
and  find  the  proceeds : 

6.  Note  of  $650,  given  Jan.  8,  1897,  payable  60  da. 
after  date;  discounted  Feb.  1,  at  1%  per  month. 

7.  Note  of  $1840,  given  July  5,  1897,  payable  in 
30  da. ;  discounted  July  8,  at  1  %  per  month. 

8.  Note  of  $2550,  given  May  3,  1897,  payable  in 
3  mo.,  with  interest  at  6%  ;  discounted  May  3,  at  6%. 

Note.  The  note  bears  interest  for  3  mo.  3  da. ;  but  the  term 
of  discount  is  95  da.,  which  is  3  mo.  5  da.,  as  banks  reckon  time. 

9.  Note  of  $56.25,  given  July  29,  1897,  payable  in 
6  mo.,  with  interest  at  10%  ;  discounted  Oct.  1,  at  1% 
per  month. 

ANNUAL  INTEREST 

55.  If  a  note  reads  "  with  interest  payable  annually,'^ 
or  "  with  annual  interest,"  the  interest  is  due  at  the  end 
of  each  year,  and  thereafter  draws  simple  interest  until 
paid.     Interest  so  computed  is  called  annual  interest. 

Ex.  1. 

$  1000.00.  Chicago,  Jan.  13,  1897. 

Three  and  one-half  years  from  date,  I  promise  to  pay 
Henry  Ames,  or  bearer.  One  Thousand  Dollars,  for  value 
received,  with  interest  at  6%,  payable  annually. 

M.  J.  Clarkson. 


76  ALGEBRAIC   ARITHMETIC 

Find  the  amount  of  the  note  at  maturity  (not  counting 
days  of  grace),  no  interest  having  been  paid. 

The  simple  interest  on  the  principal  for  3^  yr.  =  $  210. 
The  annual  interest  on  the  principal  =    $  60. 

The  first  year's  interest  remains  unpaid  2^  yr. ;  the 
second,  li  yr. ;  the  third,  i  yr.  This  is  equivalent  to 
the  use  of  $  60  for  (2^  +  1 1-  +  i)  yr.,  or  4^  yr. 

Interest  on  $  60  for  ^  yr.  =  $  16.20. 

Total  interest  =  $  210  +  1 16.20  =  $  226.20. 

Amount  due  at  maturity = $  1000 + $  226.20 = $  1226.20. 

EXAMPLES  29 

Find  the  annual  interest  of : 

2.  $8000for  5yr.,  at6%. 

3.  $  1500  for  4  yr.,  at  7%. 

4.  $3500  for  10  yr.,  at  8%. 

5.  $675for9iyr.,  at8%. 

6.  $800  for  4  yr.,  at  7%. 

COMPOUND  INTEREST 

56.  In  compound  interest,  the  interest,  when  due,  is 
added  to  the  principal,  thus  forming  a  new  principal 
for  the  next  period.  The  interest  may  be  compounded 
with  the  principal  annually,  semiannually,  or  quarterly, 
according  to  agreement.  Annual  periods  are  understood 
unless  otherwise  stated. 


COMPOUND   INTEREST  77 

Ex.  1.   What  is  the  compound  interest  of  $750  for 
1  yr.  3  mo.,  at  8%  per  annum,  payable  semiannually  ? 

$  750  1st  principal. 
1.04 

$  780  amount  at  end  of  J  yr.,  2d  principal. 
1.04 


f  811.20  amount  at  end  of  1  yr.,  3d  principal. 
1.02 


$  827.42  amount  in  1  yr.  3  mo. 
750.00 


$  77.42  compound  interest. 

EXAMPLES  30 

2.  Find  the  simple  interest,  annual  interest,  and  com- 
pound interest  of  $2500  for  6  yr.,  at  6%. 

3.  Find  the  amount  of  $350  in  3  yr.,  at  7%  com- 
pound interest. 

Find  the  compound  interest  on : 

4.  $  1200,  for  3  yr.,  at  5%,  payable  annually. 

5.  $864.50,  for  4  yr.,  at  8%,  payable  annually. 

6.  $  680,  for  2  yr.,  at  7%,  payable  semiannually. 

7.  $460,  for  1  yr.  5  mo.  18  da.,  at  6  %,  payable  quar- 
terly. 

8.  $  1250,  for  3  yr.  7  mo.  18  da.,  at  5  %,  payable  semi- 
annually. 

9.  $  790,  for  9  mo.  27  da.,  at  8%,  payable  quarterly. 


78  ALGEBRAIC   ARITHMETIC 

10.  What  sum  placed  at  simple  interest  for  3  yr.  10  mo. 
18  da.,  at  7  %,  will  amount  to  the  same  as  $  1500  placed 
at  compound  interest  for  the  same  time,  and  at  the 
same  rate,  payable  semiannually  ? 

11.  How  much  must  a  father,  at  the  birth  of  his  son, 
set  apart  for  his  benefit,  so  that  with  the  interest  at  7%, 
compounded  semiannually,  it  may  amount  to  $10,000 
when  his  son  shall  become  21  years  of  age  ? 

PARTIAL  PAYMENTS 

57.  When  partial  payments  are  made  on  notes  or  other 
obligations  bearing  interest,  they  may  be  applied  in  either 
of  two  ways ;  namely : 

(1)  To  the  debt  of  principal,  leaving  the  interest  un- 
paid till  the  time  of  final  settlement. 

(2)  To  the  debt  of  interest  first,  and  the  remainder,  if 
any,  to  the  principal. 

There  are  other  methods  which  are  formed  by  various 
combinations  of  these  two. 

It  will  be  seen  from  the  following  articles  that  by  the 
first  method  the  debt  draws  simple  interest ;  by  the  sec- 
ond, compound  interest. 

Note.  All  interest  is  in  effect  compounded  when  it  is  paid, 
since  it  allows  the  lender  to  loan  it  again  and  so  draw  interest  on 
interest,  while,  If  not  paid,  the  debtor  has  the  use  of  the  interest 
money  without  paying  interest  on  it. 

The  acknowledgment  of  a  partial  payment,  stating 
the  time  and  amount  of  the  same,  is  written  on  the  back 
of  the  note  j  and  is  called  an  indorsement. 


PARTIAL  PAYMENTS  79 

58.  The  first  method  mentioned  in  the  last  article  is 
commonly  used  when  partial  payments  are  made  on  mer- 
cantile accounts  which  are  past  due,  and  on  notes  con- 
taining the  words  "  with  interest "  and  running  for  a  year 
or  less.     It  is  called  the  Merchants'  Rule. 

Ex.  1.  What  is  due  Oct.  1,  1897,  on  a  note  of  $  750, 
with  interest  at  6%,  dated  June  1,  1897,  and  bearing  the 
following  indorsements :   July  1,  ^  100 ;  Aug.  19,  $  250  ? 

Interest  on  1st  principal  for  1  mo.  =$  750  x  .005  =  $  3.75 

Interest  on  2d  principal  for  1  mo.  18  da.    =    650  x  .008  =      5.20 
Interest  on  3d  principal  for  1  mo.  12  da.    =    400  x  .007  =      2.80 

Total  interest $11.75 

Amount  due  Oct.  1  =  $400  +  $11.75  =  $411.75. 

The  same  result  is  obtained  by  the  following  method 
of  procedure,  which  is  the  usual  way  of  stating 


The  Merchants'  Rule 

Find  the  amount  of  the  note  or  debt  from  its  date  to  the 
time  of  settlement. 

Find  the  amount  of  each  payment  from  its  date  to  the 
time  of  settlement. 

Subtract  the  sum  of  the  amounts  of  the  payments  from 
the  amount  of  the  note  or  debt. 

Thus,  in  the  above  example, 
Amount  of  $  750  for  4  mo.  is  .     .     .  $  765.00 

Amount  of  $  100  for  3  mo.  is  .     .     .     $  101.50 
Amount  of  $  250  for  1  mo.  12  da.  is        251.75        353.25 

Amount  due S  411.75 


80  ALGEBRAIC   ARITHMETIC 

EXAMPLES   31 

Write  out  the  following  in  proper  form  on  paper, 
placing  the  indorsements  on  the  back,  and  solve  by  the 
Merchants^  Eule : 

2.  Face,  ^1500.  Date,  Jan.  1,  1895.  Interest,  6%. 
Indorsements:  Aug.  7,  1895,  $500;  Dec.  7,  1895,  $500. 
What  is  due  Jan.  1,  1896? 

3.  Face,  $480.  Date,  March  3,  1894.  Interest,  7%. 
Indorsements:  Sept.  3,  1894,  $196.80;  March  3,  1895, 
$  214.     Sept.  3,  1895,  paid  the  amount  due.     Find  it. 

4.  Face,  $1000.  Date,  July  20,  1894.  Interest,  8%. 
Indorsements  :  March  5, 1895,  $  50 ;  July  5,  1895,  $  450. 
What  was  still  due  on  the  date  of  last  payment  ? 

5.  Face,  $1230.  Date,  Jan.  1,  1896.  Interest,  51%. 
Indorsements  :  March  1,  1896,  $  98 ;  June  7, 1896,  $  500 ; 
Sept.  20,  1896,  $290;  Dec.  10,  1896,  $100.  What  is 
due  Jan.  1,  1897  ? 

6.  Face,  $800.  Date,  March  1,  1896.  Interest,  10%. 
Indorsements:  Aug.  10,  1896,  $200;  Sept.  1,  1896,  $50; 
Jan.  1,  1897,  $  15.     What  was  due  March  1,  1897  ? 

59.  The  second  method  of  applying  partial  payments, 
mentioned  in  Art.  57,  is  generally  employed  in  the  case 
of  interest-bearing  notes  that  run  for  more  than  a  year ; 
but  is  also  frequently  used  when  the  time  is  less  than  a 
year. 

Under  the  application  of  this  method,  three  cases  may 
arise ;  namely,  a  payment  may  be  (1)  equal  to,  (2)  greater 
than,  or  (3)  less  than,  the  interest  accumulated  at  the 
time  of  the  payment. 


PARTIAL  PAYMENTS  81 

In  the  first  case,  the  payment  just  cancels  the  interest, 
and  the  principal,  or  interest-bearing  debt,  remains  un- 
changed. The  debtor,  in  the  end,  pays  just  as  much  as 
if  such  payment  had  been  deferred  until  he  was  able  to 
make  a  payment  large  enough  to  diminish  the  principal; 
and,  meanwhile,  he  loses  the  use  of  the  payment. 

In  the  second  case,  the  principal  is  diminished ;  hence 
the  total  interest  on  the  debt  is  diminished  by  such 
payment. 

In  the  third  case,  if  the  unpaid  balance  of  the  interest 
were  added  to  the  principal,  the  interest-bearing  debt 
would  be  increased.  This  would  increase  the  total  in- 
terest ;  and  besides  losing  the  use  of  such  payment,  the 
debtor  would  actually  have  more  to  pay,  in  the  end,  than 
if  he  had  kept  the  money  till  he  was  able  to  make  a 
sufficiently  large  payment  to  reduce  the  principal. 

It  was  to  prevent  such  manifest  injustice  to  the  debtor 
that  the  Supreme  Court  of  the  United  States  adopted  the 
following  rule : 

The  United  States  Rule 

Find  the  amount  of  the  principal  to  the  time  when  the 
payment,  or  the  sum  of  the  payments,  equals  or  exceeds 
the  interest. 

From  this  amount  deduct  the  payment  or  sum  of  the 
payments. 

Consider  the  remainder  as  a  new  principal,  and  proceed 
as  before. 

Ex.  1.   A  note  of  $500,  dated  Feb.  1,  1895,  and  bear- 
ing interest  at  6%,  is  indorsed  as  follows:  May  1,  1895, 
o 


82  ALGEBRAIC   ARITHMETIC 

^40;  Nov.  14,  1895,  ^8;  April  1,  1896,  $18;   May  1, 
1896,  $30.    What  was  due  Sept.  16, 1896  ? 

$  500.00  1st  principal. 

7.50  interest  to  May  1. 
$507.50 

40.00  1st  payment. 
$  467.50  2d  principal. 

15.04  interest  to  Nov.  14. 

10.67  interest  to  April  1. 
$493.21 

26.00  2d  and  3d  payments. 
$467.21  3d  principal. 

2.34  interest  to  May  4. 
$  469.55 

30.00  4tli  payment. 


$  439.55  4tli  principal. 

9.89  interest  to  Sept.  16. 
$  449.44  amount  due. 

Note.  It  will  be  seen  from  the  note  to  Art.  57  that  by  the 
United  States  Rule  the  interest  is  compounded  as  often  as  a  pay- 
ment is  made  which  equals  or  exceeds  the  unpaid  interest. 

EXAMPLES  32 

2  to  6  inclusive.  Solve  Ex.  2  to  6  inclusive  of  the  last 
article  by  the  United  States  Rule,  and  compare  the  results 
with  those  obtained  by  the  Merchants'  Rule.  Account 
for  the  difference  in  the  results. 

7.  What  was  due  Aug.  5,  1896,  on  a  note  for  $  2500, 
with  interest  at  7%,  dated  Aug.  5, 1895,  and  bearing  the 


PARTIAL  PAYMENTS  83 

following  indorsements :  Jan.  1,  1896,  $  500 ;  March  10, 

1896,  $  750  ? 

8.  A  note  for  $  16,500,  dated  May  20,  1896,  and  bear- 
ing interest  at  7%,  is  indorsed  as  follows :  Sept.  1,  1896, 
$  25 ;  Oct.  14, 1896,  $  150 ;  March  20, 1897,  $  45 ;  July  5, 

1897,  $  300.     Find  the  amount  due  Nov.  11,  1897. 

9.  Find  the  amount  due  Jan.  1,  1897,  on  a  note  for 
$497.39,  with  interest  at  6%,  dated  March  15,  1894,  and 
indorsed  as  follows:  Nov.  3,  1894,  $57.50;  June  15, 
1895,  $  22.25 ;  Aug.  1, 1895,  $  125 ;  Sept.  15,  1895,  $  175. 

10.  A  note  for  f  10,000  runs  4  yr.,  at  8%  interest,  on 
which  were  made  quarterly  payments  of  $500.  What 
was  the  amount  due  at  the  time  of  settlement  ? 

11.  On  a  note  for  $1000,  at  6%  interest,  payments 
were  made  as  follows:  in  1  yr.,  $50;  in  1  yr.  6  mo., 
$250;  in  2  yr.,  $224;  in  2  yr.  8  mo.,  $20;  in  2  yr. 
10  mo.,  $  110.    Find  the  amount  due  at  the  end  of  4  yr. 


CHAPTER  V 

PROPORTION.    PARTNERSHIP.    AVERAGE  OF 
PAYMENTS 

60.  Ratio.  The  relative  magnitude  of  two  numbers, 
measured  by  the  quotient  of  the  first  divided  by  the 
second,  is  called  their  ratio. 

Thus  the  ratio  of  12  to  3  is  4 ;  of  9  da.  to  4  da.  is  2\ ; 
of  3  pt.  to  1  gal.,  is  f . 

Concrete  numbers  of  different  kinds  can  have  no  ratio 
to  one  another.  For  example,  we  cannot  compare  feet 
and  pounds  with  respect  to  their  magnitude.  Moreover, 
concrete  numbers  of  the  same  kind  must  be  expressed 
in  the  same  unit  before  their  ratio  can  be  taken. 

A  ratio  is  always  an  abstract  number,  and  may  be  ex- 
pressed as  a  rate  per  cent  (Case  II,  Art.  32).  Thus,  in 
percentage,  the  rate  is  the  ratio  of  the  percentage  to 
the  base. 

The  ratio  of  any  two  numbers  a  and  b  is  expressed 

by  the  notation  a :  6  or  - ;  and  a  is  called  the  first  term 

of  the  ratio,  or  the  antecedent;  and  6,  the  second  term, 
or  the  consequent. 

The  product  of  two  or  more  ratios  is  called  a  compound 
ratio. 

Thus  the  ratio  compounded  of  the  ratios  3  :  4  and  5  :  7 
is  16  :  28 ;  since  J  X  ^^  =  if . 

84 


PROPORTION  85 


PROPORTION 

61.  A  statement  of  the  equality  of  two  ratios  is  called 
a  proportion,  and  is  expressed  in  three  ways ;  thus : 


t  =  A, 

4 

:  8  =  6 :  12, 

4 

:  8  : :  6  :  12. 

The  last  is  the  usual  notation,  and  is  read  "4  is  to  8 
as  6  is  to  12." 

The  four  terms  of  a  proportion  are  said  to  be  propor- 
tional or  in  proportion. 

Thus  4,  8,  6,  12,  are  proportional. 

The  first  ratio  of  a  proportion  is  called  the  first  couplet ; 
the  second  ratio,  the  second  couplet. 

The  first  and  fourth  terms  of  a  proportion  are  called 
extremes ;  and  the  second  and  third  terms,  means. 

62.  Denote  any  four  proportional  numbers  by  the 
letters  a,  6,  c,  d;  then 

a:b::  c:d,  (1) 

Multiply  the  sides  of  this  equation  by  bd ;  then 

ad=bc.  (3) 

Hence  the  law : 

(i.)  The  product  of  the  extremes  of  any  proportion  is 
equal  to  the  product  of  the  means. 


86  ALGEBBAIC   ARITHMETIC 

If  equation  (2)  or  (3)  be  solved  for  each  of  the  quan- 
tities in  succession,  we  obtain 

6c     ,     be  x.s 

a  =  -,  (^=-;  (4) 

a  a 

,      ad    ^     ad  .f.. 

6  =  —,  c  =  — .  (5) 

c  0 

From  (4)  we  have  the  law : 

(ii.)  The  product  of  the  means  divided  by  either  extreme 
will  give  the  other  extreme. 

From  (5)  we  have  the  law : 

(iii.)  The  product  of  the  extremes  divided  by  either  mean 
will  give  the  other  mean. 

From  (i.)  it  follows  that  a  proportion  is  verified,  or 
proved,  by  showing  that  the  product  of  the  extremes  is 
equal  to  the  product  of  the  means. 

From  (ii.)  and  (iii.)  it  follows  that  if  three  terms  of  a 
proportion  are  given,  the  fourth  term  can  be  found. 

EXAMPLES  33 

Verify  the  following  proportions : 

1.  12  :  1728  :  :  1 :  144. 

2.  27.03  :  9.01 ::  16.05 :  5.35. 

3.  i:*::f:A. 

Find  the  value  of  x  in  each  of  the  following  propor- 
tions : 

4.  8:  62::  20:  a;.  6.    «:  20  : :  120  :  50. 
6.   12:a;::l:144.                 7.   80:4::a;:i. 


PROPORTION  87 

8.  2.5:  62.5::  5:  a;.  12.    | :  a; :  :  J  :  59.0625. 

9.  175.35:a;::i:f  13.   tV^I^-^^-I- 

10.  4|:a;::9f  :27^.  14.   a; :  38J  :  :  8| :  76^. 

11.  ic  :  9.01 ::  16.05  :  5.35.       15.    7.5  :  18  :  :  a; :  7tV 

63.  If  four  numbers  a,  h,  c,  c?,  are  proportional,  that 

is,  if 

a  :  6  : :  c :  d,  (1) 

then  it  is  also  true  that 

6  :  a  : :  d  :  c,  (2) 

a.ci'.hidj  (3) 

c'.a.'.d'.h.  (4) 

This  is  easily  proved ;  for,  from  (1),  we  know  that 

ad  =  be,  (Art.  62.)  (5) 

or  be  =  ad.  (6) 

Divide  the  sides  of  (6)  by  ac ;  then 

-  =  - ,  or  b:  a:  :d:  c. 
a      c 

Similarly,  dividing  the  sides  of  (6)  by  ab  gives  (4); 
and  dividing  the  sides  of  (5)  by  cd  gives  (3). 

Four  other  proportions  can  be  obtained  from  the  four 
given  by  interchanging  the  couplets. 

Exercise.  Express  the  proportionality  of  the  num- 
bers 5,  15,  7,  21,  in  as  many  ways  as  possible. 

64.  When  any  substance  is  sold  at  a  fixed  price  per 
pound,  the  cost  of  any  amount  of  it  is  so  related  to  its 


88  ALGEBRAIC   ARITHMETIC 

weight  that  when  the  weight  is  doubled  the  cost  is  also 
doubled,  when  the  weight  is  halved  the  cost  is  also 
halved,  and  so  on.  This  relation  is  expressed  by  saying 
that  the  cost  and  the  weight  are  proportional. 

For  example,  if  coffee  is  20  cents  per  lb.,  3  lb.  will 
cost  60  cents,  and  5  lb.  will  cost  100  cents ;  and 

3  lb.  ^  20  ct. 

61b.     100  ct.' 

or  3  lb. :  5  lb. :  :  20  ct. :  100  ct. 

And,  in  general,  if  a  pounds  of  the  coffee  cost  f  h  and 
c  pounds  of  it  cost  $  d,  then 

alb.  :clb.  ::$6:$c?; 

which  is  the  symbolical  statement  of  the  fact  that  the 
ratio  of  any  two  values  of  the  cost  is  equal  to  the  ratio 
of  the  corresponding  weights. 

Definition.  One  quantity  is  said  to  be  proportional 
to  another  when  the  two  are  so  related  that  the  ratio  of 
any  two  values  of  the  one  is  equal  to  the  ratio  of  the 
corresponding  values  of  the  other. 

It  should  be  observed  that  in  the  proportion,  corre- 
sponding values  are  both  antecedents  or  both  consequents. 

65.  Definition.  The  reciprocal  of  a  number  is  1  di- 
vided by  that  number.  Hence  the  reciprocal  of  a  fraction 
is  the  fraction  inverted,  and  the  reciprocal  of  a  ratio  is 
the  ratio  formed  by  interchanging  its  antecedent  and 
consequent. 

If  the  number  of  men  to  do  a  given  piece  of  work  be 
doubled,  the  time  required  to  do  the  work  will  be  halved  j 


PROBLEMS  89 

if  3  times  as  many  men  work,  the  time  required  will  be 
^  as  long ;  if  only  \  as  many  men  work,  it  will  take  4 
times  as  long;  and  so  on.  This  relation  is  expressed 
by  saying  that  the  time  required  to  do  the  work  is  in- 
versely proportional  to  the  number  of  men  working. 

For  example,  if  1  man  can  do  a  piece  of  work  in  48  da., 
4  men  can  do  it  in  J  of  48  da.,  or  12  da.,  and  6  men  can 
do  it  in  ^  of  48  da.,  or  8  da. 

.  ^  4  men      8  da. 

Also, =  — -— — t 

'  6  men     12  da. 

or  4  men :  6  men  : :  8  da. :  12  da. ; 

from  which  it  will  be  seen  that  the  ratio  of  the  two  num- 
bers of  men  is  equal  to  the  reciprocal  of  the  ratio  of  the 
corresponding  numbers  of  days. 

Definition.  One  quantity  is  said  to  be  inversely 
proportional  to  another  when  the  ratio  of  any  two  values 
of  the  one  is  equal  to  the  reciprocal  of  the  ratio  of  the 
corresponding  values  of  the  other. 

Observe  that  in  the  case  of  inverse  proportionality, 
corresponding  values  are  both  extremes  or  both  means. 

PROBLEMS  IN  SIMPLE  PROPORTION 

66.  A  statement  of  the  equality  of  two  simple  ratios 
is  called  a  simple  proportion. 

Problems  involving  two  pairs  of  quantities,  propor- 
tional or  inversely  proportional,  three  of  which  quantities 
are  given,  can  be  solved  by  simple  proportion. 

Ex.  1.  If  20  lb.  of  sugar  cost  $1.20,  what  will  45  lb. 
cost  ? 


90  ALGEBRAIC   AEITHMETIC 

Let  $  X  denote  the  cost  of  45  lb.  of  sugar  j  then,  sincej^  T 
the  cost  is  proportional  to  the  weight,  \   ^  i^ 

201b.  :451b.::  $1.20  :$a;.  ^*     i     "^N, 

Hence     $  a;  =  ^^  x  $  1.20  =  $  2.70.  ^ns. 
201b. 

Note.  In  concrete  problems  the  product  of  the  extremes  or  of 
the  means  will  be  the  product  of  two  concrete  numbers,  and  this 
has  no  meaning.  A  multiplier  is  necessarily  an  abstract  number. 
Hence  in  the  above  example  we  cannot  multiply  $  1.20  by  45  lb. ; 
but  we  can  multiply  it  by  the  ratio  of  45  lb.  to  20  lb. ,  for  all  ratios 
are  abstract  numbers  (Art.  60). 

Since  the  ratio  of  45  lb.  to  20  lb.  is  the  same  as  the  ratio  of  45 
to  20,  it  is  unnecessary  to  retain  concrete  denominations  in  the 
solution,  except  in  the  case  of  the  number  of  the  same  kind  as 
the  answer.    Hence  we  may  proceed  as  follows : 

.06 

The  required  quantity  may  be  taken  as  any  one  of  the  four 
terms  of  the  proportion,  but  it  is  customary  to  write  it  as  the 
fourth. 

Solution  by  Analysis.  If  20  lb.  of  sugar  cost  $1.20,  1  lb. 
will  cost  -^^  of  $1.20,  or  6^,  and  45  lb.  will  cost  45  times  6)*,  or 
$2.70. 

Ex.  2.  In  how  many  days  can  12  men  do  a  piece  of 
work  that  60  men  can  do  in  8  da.  ? 

The  number  of  men  and  the  number  of  days  are  in- 
versely proportional;   hence 

12  men :  60  men  : :  8  da. :  a:  da. 


PROBLEMS  91 

Solution  by  Analysis.     Since  it  takes  60  men  8  da.,  it  will 
take  1  man  60  times  8  da.,  and  it  will  take  12  men  ^^  of  60  x  8  da., 
Q^  60x8  da. 
12 

EXAMPLES  34 

Note.  All  problems  in  proportion  can  be  solved  by  analysis. 
The  learner  should  become  familiar  with  both  methods  of  solution. 

3.  If  20  yd.  of  cloth  cost  $  180,  find  the  cost  of  45  yd. 

4.  If  18  bu.  of  wheat  make  4  bbl.  of  flour,  how  many 
barrels  will  200  bu.  make  ? 

5.  How  many  men  will  be  required  to  build  32  rods 
of  wall  in  the  same  time  that  5  men  can  build  10  rods  ? 

6.  If  5  sheep  can  be  bought  for  ^  20.75,  how  many 
sheep  can  be  bought  for  $  398.40  ? 

7.  When  10  bbl.  of  flour  cost  $  112.50,  what  will  be 
the  cost  of  476  bbl.  of  flour  ? 

8.  If  a  train  runs  30  mi.  in  50  min.,  in  what  time  will 
it  run  260  mi.  ? 

9.  If  a  horse  travels  12  mi.  in  1  hr.  36  min.,  how  far 
at  the  same  rate  will  he  travel  in  15  hr.  ? 

10.  How  many  days  will  12  men  require  to  do  a  piece 
of  work  that  95  men  can  do  in  7^  da.  ? 

11.  If  f  of  an  acre  of  land  cost  $60,  what  will  45f 
acres  cost? 

12.  If  by  selling  $5000  worth  of  dry  goods  a  mer- 
chant gains  $  456.25,  what  amount  must  he  sell  to  gain 
$  1000  ? 


92  ALGEBRAIC   ARITHMETIC 

13.  If  a  pasture  will  feed  120  horses  81  da.,  how  many- 
horses  will  it  feed  108  da.  ? 

14.  If  a  business  yields  $  700  profits  in  1  yr.  8  mo.,  in 
what  time  will  it  yield  f  1050  profits  at  the  same  rate  ? 

15.  If  it  takes  a  train  2^  hr.  to  go  a  certain  distance  at 
the  rate  of  27  mi.  an  hour,  how  long  will  it  take  to  go  the 
same  distance  at  the  rate  of  21  mi.  an  hour  ? 

16.  If  15  men  can  build  a  wall  in  6  da.,  how  many 
men  would  be  required  to  build  it  in  4^  da.? 

17.  If  a  quantity  of  provisions  is  sufficient  to  support 
225  men  25  da.,  how  many  days  will  it  support  75  men  ? 

18.  If  12  men  earn  $  78  in  4  da.,  how  many  men  will 
earn  $  58 J  in  the  same  time  at  the  same  wages  ? 

COMPOUND  PROPORTION 

67.  A  statement  of  the  equality  of  two  compound 
ratios,  or  of  a  compound  ratio  and  a  simple  one,  is  called 
a  compound  proportion. 

For  example,  the  equation 

may  be  expressed  as  a  compound  proportion ;  thus : 

3:4 

2 


;J}::9:30; 


in  which  form  the  product  of  the  ratios  written  one  above 
the  other  is  understood.  By  taking  the  product,  the  pro- 
portion is  reduced  to  a  simple  one. ,  Thus  the   above 

becomes 

6  :  20  : :  9  :  30. 


COMPOUND  PROPORTION  93 

From  the  meaning  of  a  compound  proportion,  it  follows 
that  the  product  of  the  extremes  is  equal  to  the  product 
of  the  means,  as  in  the  case  of  a  simple  proportion; 
hence  a  missing  term  is  found  in  the  same  way. 

Ex.  1.   Find  the  value  of  x  in  the  proportion 

3 

The  proportion  means  that 

5  X  3  :  6  X  7  : :  10  :  aj. 


"Vr-^- 


Hence  ^^6x7x10^^3 

5x3 

68.  The  amount  of  work  done  by  a  given  number  of 
men  is  proportional  to  the  time,  and  the  amount  of  work 
done  in  a  given  time  is  proportional  to  the  number  of 
men.  If  both  the  time  and  the  number  of  men  vary,  the 
amount  of  work  is  proportional  to  their  product. 

^or  example,  6  men  in  4  da.  can  do  24  times  as  much 
work  as  1  man  in  1  da. 

Hence,  if  1  man  can  dig  2  rd.  of  ditch  in  1  da.,  6  men 
in  4  da.  can  dig  6x4x2  rd.,  or  48  rd.,  and  5  men  in 
3  da.  can  dig  5  x  3  x  2  rd.,  or  30  rd. 

.,  6men     4da._48rd. 

^^''''  5  men  ^3  da.  ~  30  rd.' 

6  men:  5  men)       .^    ,     ^^    , 
or  .  ,         o  ^       ^  : :  48  rd.  :  30  rd. ; 

4  da.    :  3  da.    ) 

that  is,  the  ratio  of  the  two  amounts  of  work  is  equal  to 
the  product  of  the  ratios  of  the  corresponding  numbers 
of  men  and  days. 


94  ALGEBRAIC   ARITHMETIC 

Definition.  One  quantity  is  said  to  be  proportional 
to  the  product  of  two  or  more  other  quantities  when  the 
ratio  of  any  two  values  of  that  quantity  is  equal  to  the 
product  of  the  ratios  of  the  corresponding  values  of 
the  others. 

PROBLEMS  IN  COMPOUND  PROPORTION 

69.  Ex.  1.   If  18  men  build  126  rd.  of  wall  in  60  da., 
how  many  rods  will  6  men  build  in  110  da.  ? 
Symbolical  statement : 

18  men  :  6  men    ]      ^  o^    i  i 

60  da.    :110da.r  =  ^^^^^-  =  ^^^- 

Hence    g.  rd.  =  ^^  ><  ,^f  xj^^  ^^- =  77  rd. 
18  X  60 

Solution  by  Analysis.    One  man  in  60  da.  will  build  —  of 

126 
126  rd.,  or  —  rd. :  6  men  in  the  same  time  will  build  6  times  as 
18 

many  rods,  or —  rd. ;  in  1  da.  the  6  men  will  build  —  of 

''is  60 

126ji6  ^         126x_6  j,^     ^^^  .j^  j^Q  ^^    ^^       ^.j^  ^^.j^  j^Q 

18  18  X  60        ' 

times  as  many  rods  as  in  1  da.,  or  — -^ — ^-- —  rd. 

18  X  60 

Ex.  2.  If  18  men  build  126  rd.  of  wall  in  60  da.,  how 
many  men  will  it  take  to  build  77  rd.  in  110  da.  ? 

126  rd. :  77  rd. )      .^ 

■fiA  J      ar\A     > : :  18  men  :  a; men. 

110  da. :  60  da.  ) 

xj«„««    r.  r«^«      77  X  60  X  18  men      ^  _^^ 

Hence    x  men  = — — — --— =  b  men. 

126  X  110 


PROBLEMS  95 

Explanation  op  the  Method.  In  problems  of  this  class  all 
the  numbers  occur  in  like  pairs,  except  one  which  is  of  the  same 
kind  as  the  answer.  Take  this  as  the  third  term  of  the  proportion  ; 
then  the  fourth  term,  when  found,  is  the  answer.  Consider  each 
of  the  pairs  of  numbers  separately,  forming  a  first  couplet  from 
each,  as  in  simple  proportion. 

In  solving  by  analysis,  begin  with  the  number  like  the  answer,  and 
consider  the  effect  upon  it  of  the  given  change  in  each  of  the  other 
numbers,  separately.  Thus,  in  the  analysis  of  Ex.  ] ,  we  considered 
the  effect  upon  the  number  of  rods  caused  first  by  the  change  in 
the  number  of  men  from  18  to  6  (the  time  remaining  unchanged), 
then  by  the  change  in  the  number  of  days  from  60  to  110  (the  num- 
ber of  men  remaining  unchanged).  In  each  case  we  first  reason 
to  1  of  the  number  that  is  changed. 

EXAMPLES  35 

3.  If  8  men  earn  $  320  in  8  da.,  how  much  will  12  men 
earn  in  4  da.  ? 

4.  If  it  costs  $  41.25  to  pave  a  sidewalk  5  ft.  wide  and 
75  ft.  long,  what  will  it  cost  to  pave  a  similar  walk  8  ft. 
wide  and  566  ft.  long  ? 

5.  If  16  horses  consume  48  bu.  of  oats  in  12  da.,  how 
many  bushels  will  20  horses  consume  in  8  wk.  ? 

6.  What  sum  of  money  will  gain  $300  in  8  mo.,  if 
$  800  gain  |  70  in  15  mo.  ? 

7.  If  10  men  can  cut  46  cords  of  wood  in  18  da.,  work- 
ing 10  hr.  a  day,  how  many  cords  can  40  men  cut  in 
24  da.,  working  9  hr.  a  day  ? 

8.  What  is  the  cost  of  36^  yd.  of  cloth  IJ  yd.  wide,  if 
^  yd.,  If  yd.  wide,  cost  $  3.37^  ? 


96  ALGEBRAIC   ARITHMETIC 

9.  A  contractor  employs  45  men  to  complete  a  work 
in  3  mo.  What  additional  number  of  men  must  lie 
employ  to  complete  the  work  in  2^  mo.  ? 

10.  How  many  days  will  21  men  require  to  dig  a  ditch 
80  ft.  long,  3  ft.  wide,  and  8  ft.  deep,  if  7  men  can  dig  a 
ditch  60  ft.  long,  8  ft.  wide,  and  6  ft.  deep  in  12  da.  ? 

11.  When  the  shadow  of  a  post  10  ft.  6  in.  high  is 
12  ft.  3  in.  long,  what  is  the  length  of  shadow  of  a  post 
8  ft.  9  in.  high  ? 

12.  The  shadow  of  a  post  16  ft.  3  in.  high  is  5  ft.  5  in. 
long.  What  height  of  post  will  give  a  shadow  3  ft.  4  in. 
long? 

13.  If  a  vat  16  ft.  long,  7  ft.  wide,  and  15  ft.  deep  holds 
384  bbl.,  how  many  barrels  will  a  vat  17^  ft.  long,  lOJ  ft. 
wide,  and  13  ft.  deep  hold  ? 

14.  What  is  the  weight  of  a  block  of  granite  8  ft.  long, 

4  ft.  wide,  and  10  in.  thick,  if  a  similar  block  10  ft.  long, 

5  ft.  wide,  and  16  in.  thick  weighs  5200  lb.  ? 

15.  If  it  costs  $15  to  carry  20  tons  IJ  mi.,  what  will 
it  cost  at  the  same  rate  to  carry  400  tons  ^  mi.  ? 

16.  If  6  laborers  can  dig  a  ditch  34  yd.  long  in  10  da., 
how  many  days  will  20  laborers  require  to  dig  a  similar 
ditch  170  yd.  long? 

17.  If  a  man  walk  192  mi.  in  6  da.,  walking  8  hr.  a 
day,  how  far  can  he  walk  in  18  da.,  walking  6  hr.  a  day  ? 


PARTNERSHIP  97 


PARTNERSHIP 


70.  The  association  of  two  or  more  persons  for  the 
purpose  of  carrying  on  business  is  called  partnership. 

The  persons  thus  associated  are  called  partners,  and 
together  they  form  a  company  or  firm. 

The  money  or  property  invested  is  called  the  capital  or 
stock. 

The  money  and  property  of  all  kinds  belonging  to  a 
company,  including  the  amounts  due  it,  are  called  its 
resources  or  assets ;  its  debts  are  called  liabilities. 

The  profits  and  losses  of  a  company  are  usually  divided 
among  the  partners  proportionally  to  the  capital  of  each, 
if  all  invest  for  the  same  time ;  and  proportionally  to  the 
product  of  capital  and  time  if  the  times  are  different. 

Problems  in  partnership  are  therefore  solved  by  the 
same  methods  as  other  problems  in  proportion. 

EXAMPLES  36 

1.  A  and  B  form  a  partnership.  A  furnishes  $400 
capital  and  B  $600.  They  gain  $250.  What  is  the 
profit  of  each  ? 

Suggestion.     (1)   The  whole  capital  is  ^  1000 ;  hence 
$  1000  :  $  400  : :  $  250  :  A's  share. 
$  1000  :  $  600  : :  $  250  :  B's  share. 

(2)  Each  partner  receives  the  same  fraction  (or  per  cent)  of  the 
whole  gain  that  his  capital  is  of  the  whole  capital. 

(3)  The  gain  of  each  partner  is  the  same  fraction  (or  per  cent) 
of  his  capital  that  the  whole  gain  is  of  the  whole  capital. 

2.  A,  B,  and  C  traded  in  company.  A  put  in  $  8000 ; 
B,  $4500;  and  C,  $3500.  Their  profits  were  $6400. 
What  is  each  partner's  share  of  the  profits  ? 


98  ALGEBRAIC   ARITHMETIC 

3.  A  and  B,  in  trading  for  3  yr.,  make  a  profit  of 
$4800.  A  invested  |  as  much  stock  as  B.  What  is 
each  man's  share  of  the  profits  ? 

4.  Brooks  &  Co.  fail  in  business;  their  liabilities 
amount  to  $22,000;  their  resources  to  $8800.  Tliey 
owe  A  $4275,  and  B  $2175.50.  What  will  each  of 
these  creditors  receive? 

5.  Four  persons  engage  in  manufacturing,  and  invest 
jointly  $  22,500.  At  the  end  of  a  certain  time  A's  share 
of  the  gain  is  $2000;  B's,  $2800.75;  C's,  $1685.25; 
and  D's,  $  1014.     How  much  capital  did  each  put  in  ? 

6.  Three  partners,  A,  B,  and  C,  furnish  capital  as 
follows :  A,  $  500  for  2  mo. ;  B,  $  400  for  3  mo. ;  C, 
$  200  for  4  mo.  They  gain  $  600.  What  is  each  part- 
ner's share  ? 

Suggestion.  The  use  of  $500  for  2  mo.  is  equivalent  to  the 
use  of  !|  1000  for  1  mo.  ;  of  $  400  for  3  mo.  to  $  1200  for  1  mo.  ; 
of  $200  for  4  mo.  to  $800  for  1  mo.  Hence,  divide  the  profits 
proportionally  to  1000,  1200,  and  800. 

7.  A,  B,  and  C  gain  in  trade  $8000.  A  furnishes 
$  12,000  for  6  mo. ;  B,  $  10,000  for  8  mo.  j  and  C,  $  8000 
for  11  mo.     Apportion  the  gain. 

8.  Jan.  1,  1896,  three  persons  began  business  with 
$1300  capital  furnished  by  A.  March  1,  B  put  in 
$1000;  and  Aug.  1,  C  put  in  $900.  The  profits  at  the 
end  of  the  year  were  $  750.     Apportion  it. 

9.  In  a  certain  firm  B  has  3  times  as  much  capital  as 
A,  and  C  has  ^  as  much  as  the  other  two.  What  is  each 
one's  share  in  a  loss  of  $  786  ? 


AVERAGE   OF   PAYMENTS  99 

10.  In  a  gain  of  $  600  A  received  ^  ;  B,  ^ ;  and  C  the 
remainder.  If  the  whole  capital  was  12  times  A^s  gain, 
what  was  the  capital  of  each  ? 

11.  Two  men  receive  ^1000  for  grading.  One  fur- 
nishes 3  teams  20  da.,  and  the  other  5  teams  30  da. 
If  the  first  receives  $  100  for  overseeing  the  work,  what 
does  each  receive  ? 

12.  Two  men  contract  to  move  $  5316  cu.  yd.  of  gravel 
at  25  cents  a  cu.  yd.,  and  agree  to  share  the  profits  in  the 
ratio  of  2  to  3.  They  employ  5  teams  45  da.,  at  $4 
each  per  day.     What  did  each  make  ? 

AVERAGE  OF  PAYMENTS 

71.  Ex.  1.   A  owes  B  $  1200,  of  which  8  300  is  due  in 

4  mo.,  $  400  in  6  mo.,  and  $  500  in  12  mo.  If  he  wishes 
to  pay  the  whole  debt  at  one  time,  when  must  he  do  so  in 
order  that  neither  party  shall  lose  ? 

The  loss  that  is  here  referred  to  is  the  loss  of  the  use 
of  money,  which  is  really  loss  of  interest. 

If  A  should  pay  the  debt  at  once,  he  would  lose  the 
use  of  $300  for  4  mo.,  $400  for  6  mo.,  and  $500  for 
12  mo. ;  to  all  of  which  he  is  entitled. 

The  use  of  $  300  for  4  mo.  =  the  use  of  $  1  for  1200  mo. 
The  use  of  400  for  6  mo.  =  the  use  of  1  for  2400  mo. 
The  use  of       500  for  12  mo.  =  the  use  of     1  for  6000  mo. 


$1200  9600  mo. 

Hence  A  is  entitled  to  the  use  of  $  1  for  9600  mo.,  or 
to  the  use  of  the  $  1200  for  ^\^  of  9600  mo.,  or  8  mo. 
That  is,  the  whole  debt  will  be  due  in  a  single  payment 
in  8  mo. 


100  ALGEBRAIC   ARITHMETIC 

EXAMPLES  37 

2.  On  Dec.  1,  1896,  a  man  gave  three  notes,  the  first 
for  f  500,  payable  in  3  mo. ;  the  second  for  $  750,  payable 
in  6  mo. ;  and  the  third  for  $  1200,  payable  in  9  mo.  Find 
the  average  time  of  payment. 

3.  Bought  merchandise  Jan.  1, 1895,  as  follows :  f  350 
on  2  mo.,  $500  on  3  mo.,  $700  on  6  mo.  What  is  the 
average  time  of  payment  ? 

4.  Find  the  average  date  for  paying  three  bills,  due 
as  follows:  May  31,  |100;  June  18,  $150;  July  9, 
$  200.     (Compute  each  from  May  31.) 

5.  If  I  borrow  $  250  for  8  mo.,  how  long  should  I  lend 
$  400  to  repay  me  an  equal  interest  ? 

6.  A  person  owes  a  debt  of  $1680,  due  in  8  mo.,  of 
which  he  pays  -g^  in  3  mo.,  ^  in  5  mo.,  -g-  in  6  mo.,  and  -J-  in 
7  mo.     When  is  the  remainder  due  ? 

7.  On  a  debt  of  $  2500,  due  in  8  mo.  from  Feb.  1,  the 
following  payments  were  made :  May  1,  $  250 ;  July  1, 
$  300 ;  and  Sept.  1,  $  500.     When  is  the  balance  due  ? 

8.  Dec.  1,  1894,  purchased  goods  to  the  amount  of 
$1200,  on  the  following  terms:  25%  payable  in  cash, 
30%  in  3  mo.,  20%  in  4  mo.,  and  the  balance  in  6  mo. 
Find  the  average  time  of  payment  and  the  cash  value  of 
the  goods,  computing  discount  at  7%. 


CHAPTER   VI 
INVOLUTION  AND  EVOLUTION 

72.   Involution.     Review  Art.  22. 

The  product  of  equal  factors  is  called  a  power  of  the 
factor  thus  repeated. 

The  factor  taken  once  is  called  the  first  power;  the 
product  of  two  equal  factors  is  called  the  second  power ; 
of  three  equal  factors,  the  third  power,  and  so  on. 

The  second  power  of  a  number  is  also  called  the  square 
of  the  number,  because  it  is  equal  to  the  area  of  the 
square  the  length  of  whose  side  is  the  given  number. 
For  a  similar  reason  the  third  power  of  a  number  is 
called  its  cube. 

A  number  is  said  to  be  squared  when  its  second  power 
is  taken,  and  to  be  cubed  when  its  third  power  is  taken. 

The  process  of  taking  any  power  of  a  number  is  called 
involution. 

EXAMPLES  38  (Oral) 
Find  the  indicated  power : 


2^ 

62 

V 

1 

1002 

1.22 

22 

1 

32 

1.V 

2^ 

.72 

902 

3« 

.3» 

(1)^ 

.V 

30^ 

2* 

(t)^ 

ikf 

102 

.2* 

.012 

101 


102  ALGEBRAIC   ARITHMETIC 

73.  To  find  a  Power  of  a  Product.     Study  carefully  the 
following : 

1.  62  =  (2x3)*  =  (2x3)(2x3)  =  2x2x3x3  =  22x32 
=  4x9  =  36. 

2.  103  =  (2x5)3=(2  X  5)(2  X  5)(2  X  5)  =  2  X  2  X  2  X  5 
X  5  X  5  =  2^  X  5»=  8  X  125  =  1000. 

3.  (3  ahy  =  (3  ahh)  (Sabb)  =  3  x  3  x  aabbbb  =  9  d'b*. 

4.  (a^y  =:a^  X  a^  =  aaa  x  aaa  =  a% 
or       (a^y  =  (aaay  =  aa  x  aa  x  aa  =  a^. 

A  product  is  raised  to  any  power  by  raising  each  of  its 
factors  to  that  power. 


(a'd)* 

1 

(2cy 

74.  Evolution.  The  process  of  taking  one  of  the  equal 
factors  of  a  number  is  called  evolution.  It  is  the  inverse 
of  involution. 

One  of  the  equal  factors  of  a  number  is  called  a  root 
of  the  number.  One  of  the  two  equal  factors  of  a  num- 
ber is  called  its  square  root ;  one  of  the  three  equal  fac- 
tors, its  cube  root. 

Thus,  since  25  is  the  square  of  5,  5  is  the  square  root 
of  25 ;  since  27  is  the  cube  of  3,  3  is  the  cube  root  of  27. 


Find: 

EXAMPLES 

3  39   (Oral) 

(oc)^ 

(foa'b'y 

(iaey 

(2aby 

2(aby 

i(aey 

(5aby 

3a(aby 

fabV 

THE  SQUARE  103 

The  square  root  of  a  number  is  indicated  by  the 
radical  sign  (V)  placed  before  it ;  the  cube  root  by  -^. 

Thus  V25  =  5,  ^27  =  3. 

The  figure  placed  above  the  radical  sign  indicates 
what  root  is  to  be  taken,  and  is  called  the  index  of  the 
root.     If  no  index  is  written,  2  is  understood. 

If  an  expression  consists  of  more  than  one  term  or 
factor,  the  root  of  the  whole  is  indicated  by  placing  the 
radical  sign  before  the  expression  enclosed  in  parentheses 
or  placed  under  a  vinculum ;  otherwise  the  sign  affects 
only  the  term  or  the  factor  immediately  following. 


Thus    Vl6  +  9  =  4  +  9  =  13;     V(l^  +  9)  =  V16  +  9 

EXAMPLES  40   (Oral) 
Find: 

V49         Vl-21       ^2500       V*         12a-2V4^2 


V400        ^M         V1600       VM       V25a^+aV96* 
^125       V-81         Vl-44:        ^\         ^a'b~c 
^64         -^.001       ^1  ^ff       V9W  +  2  6*c 

75.  The  Square  of  the  Sum  of  Two  Numbers. 

Review  Arts.  18  and  19. 

We  have  learned  (Art.  18)  that  a  number  is  multiplied 
by  multiplying  each  of  its  parts,  and  that  for  this  pur- 
pose it  may  be  separated  into  parts,  or  terms,  in  any 
way. 

If  both  multiplier  and  multiplicand  consist  of  more 
than  one  term,  their  product  is  the  sum  of  the  partial 


104  ALGEBRAIC   ARITHMETIC 

products  obtained  by  multiplying  each  term  of  the  mul- 
tiplicand by  each  term  of  the  multiplier  (Art.  19). 

If  multiplier  and  multiplicand  are  equal  and  are  sepa- 
rated into  parts  in  the  same  way,  the  case  is  like  that 
of  Art.  22,  Ex.  3,  and  the  exercise.  It  was  there  shown 
that 

{a  +  bf^a'  +  Zab  +  f)'.  (1) 

Since  a  and  h  may  be  any  two  numbers,  we  have  the 
law: 

The  square  of  the  sum  of  two  numbers  equals  the  square 
of  the  first  yiumher  plus  twice  the  product  of  the  numbers 
plus  the  square  of  the  second  number. 

The  operation  of  squaring  a  number  of  two  figures  is 
simplest  when  it  is  separated  into  its  tens  and  units. 
When  it  is  so  separated,  we  shall  use  t  to  denote  the 
tens,  and  w,  the  units.     In  this  case  formula  (1)  becomes 

(i^uy=e+2iu-\-u\  (2) 

Exercise.    Express  the  meaning  of  (2)  in  words. 

76.  Illustration  of  the  Formula,  (t  +  uy  is  the  area 
of  the  square  the  length  of  whose  side  is  t-^u.  The 
square  may  be  divided  into  four  parts, 
as  shown  in  the  figure.  Comparing 
the  right  member  of  the  formula  with 
the  figure,  it  will  be  seen  that  the  first 
term  is  the  area  of  the  largest  part; 
the  second  term  is  the  area  of  the 
two  rectangles;  the  last  term  is  the 
area  of  the  small  square  that  fills  out  the  corner,  and  is 
always  the  smallest  of  the  terms. 


tu 

^ 

w« 

t 

u 

t* 

<♦* 

s 

SQUARE  ROOT  105 

Note.     The  square  of  any  number  may  be  found  by  the  for- 
mula.     For  example,  324  =  32  tens  +  4  units ;  hence  t  =  320, 

M  =  4. 

EXAMPLES  41  J  ^ 

Find  by  the  formula : 

1.   561        2.   732.        3.   2082.        4.   3161 


SQUARE  ROOT 

77.  Find  ^784:, 

The  problem  may  be  stated  thus :     Find  the  side  of 
the  square  whose  area  is  784  (sq.  in.,  say). 
Or  thus :     Find  t  and  u,  when 

t^-{-2tu  +  u^  =  7S4:. 

Begin  by  taking  the  largest  value  possible  for  t.    This 
is  easily  seen  to  be  20. 

f-\-2tu-^u^  =  7S4:sq,  in. 
Subtract         f  =  400  sq.  in. 

Eemainder     =     2tu-i-u^=:  384  sq.  in. 

Compare  with  the  figure  in  the  last  article.     What  is 
the  remainder  the  area  of  ?    The  two  rectangles  and  the 


3  i-^u -^ 


small  square  have  one  dimension,  u,  in  common.  If 
placed  as  in  the  accompanying  figure,  they  form  one  long 
rectangle  whose  dimensions  are  2  i  +  w  and  u,  and  whose 
area  therefore  is  (2t  +  u)u. 


106 


ALGEBRAIC   ARITHMETIC 


How  is  the  width  of  a  rectangle  found  if  its  length 
and  area  are  given  ?  To  find  the  width  u,  we  are  obliged 
to  use  2  tj  or  40,  as  the  length,  since  the  whole  length  is 
as  yet  unknown.  This  may  give  too  large  a  value  for  u ; 
if  so,  we  take  one  less. 

384 --2^  =  384  ^40  =  9+. 

This  is  too  large ;  for  it  gives 

(2^  +  u)u  =  (40  +  9)  X  9  =  441, 

and  there  are  only  384  sq.  in.     Hence  take  w  =  8. 

This  gives      (2 «  -f-  w)w  =  (40  +  8)  x  8  =  384. 

Hence  V'^^^  =  4^0  +  8  =  48.  Arts. 


2t 


FORMULA   FOR   SQUARE    ROOT 

f-^2tu^u^  \t  +  u 

^ 202 

~  2x20 


OPERATION 

784  I  20  +  8  =  28 


-\-u 


2t-\-u 


40 
_8 
48 


400 


384 
384 


2  Hs  called  the  trial  divisor. 

2t-\-uiB  called  the  complete  divisor. 

The  formula  for  the  square  may  be  written 

(f  +  £/)2=f2-|-(2f  +  i/)(|. 


(1) 


In  this  form,  the  first  term  of  the  coefficient  of  u  is  the 
tried  divisor.  The  whole  coefficient  is  the  complete  divisors- 
it  is  the  whole  length  of  the  addition  to  the  square  of  the 
tens. 


SQUARE   ROOT 

EXAMPLES  42 

Solve  and  prove : 

1.  V1156. 

3.    VS184. 

5.    V324. 

2.    V4225. 

4.    V841. 

6.    V9604. 

107 


78.  Let  any  integer  of  three  figures  be  separated  into 
its  hundreds,  tens,  and  units,  and  denote  these  parts 
by  the  initial  letters ;  then  the  number  will  be  denoted 
by  the  expression  h  +  t  -{-  u.     Let  us  find  its  square. 


^  + 

t   4-w 

^  + 

t    -\-u 

hu  + 

tu  +  u^ 

ht  +  t' 

tu 

h'  + 

ht         + 

hu 

h^-^2ht  +  f-{-2hu-\-2tu-\-u^ 

Hence  (h-{-t-^  u)^  =h^ -\- f+ u^ +  2ht -\-2hu -i-2  tu. 

All  we  need  observe  here  is  that  the  square  of  the 
number  contains  the  square  of  each  of  the  figures  plus 
other  terms.     This  is  true  of  any  number. 

Thus,  the  square  of  48.7 

.72=         .49 
contains  I     8^.     =     64.        plus  other  parts ; 
42   .     =16    . 


The  square  of  12.34 

.   42  = 


contains 


16 


.32    = 
21       =   04 
V  ,       =1 


09 


plus  other  parts. 


108 


ALGEBRAIC   AKITHMETIO 


It  will  be  seen  from  this  that  if  the  complete  square  of 
any  number  be  separated  into  groups  of  two  figures  each, 
commencing  at  the  decimal  point,  the  number  of  groups 
(counting  the  last  figure  to  the  left  as  a  group,  if  it 
stands  alone)  will  be  equal  to  the  number  of  figures  in 
the  root ;  and  the  square  of  each  figure  of  the  root  will  lie 
wholly  within  the  corresponding  group.^ 

79.  The  square  root  of  any  number  is  found  as  follows : 
Separate  the  number  into  groups,  as  above  directed,  and 
proceed  as  in  Art.  77,  always  regarding  the  part  of  the 
root  already  found  as  so  many  tens  with  respect  to  the 
next  figure  of  the  root. 

Ex.  1.   Find  the  square  root  of  75076. 


FULL  OPERATION 

CONTRACTED 

7'50'76 

274 

7'50'76  1  274 

4 

4 

2x20  =  40 

350 

47 

350 

7 

329 

544 

329 

47 

2176 

2x270  =  540 

2176 
2176 

2176 

4 
544 

Explanation.  The  first  trial  and  complete  divisors  are  ob- 
tained from  the  formula  precisely  as  they  would  be  if  the  given 
number  were  760.  That  is  « =  20  and  w  =  7.  For  the  second 
divisors  t  —  270  and  m  =  4. 

When  the  cipher  is  omitted  from  the  trial  divisor,  as  in  tlie  con- 
tracted operation,  omit  mentally  the  right-hand  figure  of  the  divi- 
dend in  finding  the  figure  of  the  root.  Write  the  latter,  when 
found,  in  units'  place  in  the  trial  divisor,  thus  completing  it. 


SQUARE  ROOT 


109 


9'41'.57'80  I  30.685 
9 


Ex.  2.   Extract  the  square  root  of  941.578. 

Explanation.  Complete  the  last 
group  to  the  right  by  the  addition  of  a 
cipher.  Since  there  is  a  remainder 
after  using  the  last  group,  the  root  is 
not  exact ;  but  can  be  found  to  as 
many  places  as  desired  by  annexing 
groups  of  ciphers. 

The  first  trial  divisor,  6,  is  con- 
tained 0  times  in  4,  Place  0  in  the 
root ;  and  annex  0  to  the  trial  divisor, 
and  the  next  group  to  the  dividend. 

To  find  the  square  root  of  a  fraction,  take  the  square 
root  of  its  terms  separately  if  they  are  seen  to  be  perfect 
squares ;  otherwise  it  is  best  to  reduce  to  a  decimal  first, 
as  by  so  doing  evolution  is  performed  but  once. 


606 

4157 
3636 

6128 

52180 
49024 

61365 

315600 
306825 

Vi" 


■g-^* 


TT- 


EXAMPLES  43 

3.  V13225.                9.    V196.1369.  16. 

4.  V11881.              10.    V2-251521.  16. 

5.  V994009.            11.    -y/5S.U0625.  17. 

6.  V20506.24.         12.    -^17.75.  18. 

7.  V2985.5296.       13.    V10795.21.  19. 

8.  V001225.            14.    V^f.  20. 

21.  A  square  field  contains  1,016,064  sq.  ft. 
the  length  of  each  side  ? 

22.  A  square  farm  contains  361  acres.  Find  the  length 
of  one  side. 

23.  A  field  is  208  rd.  long  and  13  rd.  wide.  What  is 
the  length  of  the  side  of  a  square  field  containing  an 
equal  area? 


V30i. 

V69f 
What  is 


110  ALGEBRAIC   ARITHMETIC 

24.  If  it  costs  ^  312  to  enclose  a  field  216  rd.  long  and 
24  rd.  wide,  what  will  it  cost  to  enclose  a  square  field  of 
equal  area  with  the  same  kind  of  fence  ? 

25.  Find  the  dimensions  of  a  rectangular  field  contain- 
ing 3200  sq.  rd.,  and  twice  as  long  as  broad. 

26.  How  many  rods  of  fence  will  enclose  a  square 
field  of  4  acres? 

27.  How  many  rods  of  fence  would  be  required  to  en- 
close a  field  of  4  acres  whose  length  is  twice  its  width  ? 

28.  What  is  the  difference  between  the  areas  of  two 
fields,  one  of  which  is  14  rd.  square  and  the  other  14 
sq.  rd.  ? 

29.  An  orchard  containing  2401  trees  has  as  many  rows 
as  there  are  trees  in  a  row.     How  many  rows  has  it  ? 

80.  The  Cube  of  the  Sum  of  Two  Numbers. 

a-j-b 

a^b^-2aW^-b^ 
a»-f2a26-|-    a6* 

a3-f.3a26-4-3a62  +  63 
Hence       {a  +  by  =  a^-\-Za^b  +  Zab^  +  b^  (1) 

Exercise  1.  State  the  formula  in  words,  calling  a 
"the  first  number"  and  b  "the  second  number." 

Solve  the  following  by  the  formula,  and  verify  by  first 
taking  the  indicated  sum,  then  cubing : 

(2  +  3)»,    (3  +  5)3,    (i2+9)»,    (20-f-5)^ 


THE   CUBE 


111 


The  solution  is  the  simplest  when  the  number  is  sepa- 
rated into  its  tens  and  units.  For  this  case  we  shall 
write  the  formula  thus: 


Exercise  2.     Express  formula  (2)  in  words. 


(2) 


81.  Illustration  of  the  Formula. 


/ 

7 

^^6 


Tt- 

- 

y 

7 

/ 

/ 


^ 


/ 


Fig.  1 


Fig.  2 


Fig.  3 


Fig.  4 


{t  -f  uf  is  the  volume  of  a  cube,  the  length  of  whose 
edge  is  t-^u.  Such  a  cube  can  be  formed  from  8  solids, 
as  follows :  A  cube  whose  edge  is  t,  and  whose  volume 
is  therefore  f  (Fig.  1);  3  rectangular  solids  covering  3 
adjacent  faces  of  the  cube  and  of  thickness  w,  the  volume 
of  each  being  tht  (Fig.  2) ;  3  rectangular  solids  filling  the 
edges,  the  volume  of  each  of  which  is  tu^  (Fig.  3);  a 
small  cube  whose  edge  is  u  and  whose  volume  is  v?,  filling 
the  corner  (Fig.  4). 

The  formula  and  figures  may  be  applied  to  any  num- 
ber, if  we  regard  it  as  being  made  up  of  tens  and  units, 
as  in  Art.  76,  note. 


EXAMPLES  44 


Solve  by  the  formula : 
1.   15^  2.   233. 


68^ 


4.   127». 


112  ALGEBRAIC   ARITHMETIC 

CUBE   ROOT 

82.  Find  ^46656. 

The  problem  is  to  find  the  edge  of  a  cube  whose  volume 
is  46656  (cu.  in.,  say),  or  to  find  t  and  u  when 

^3  _,.  3  ^2^  _|.  3  ^^2  _^  ^3  ^  4gg5g  j.^     i^  (;i^) 

Begin  by  taking  the  largest  value  possible  for  t.  This 
is  30;  hence  i^  =  27000.  Subtract  from  the  correspond- 
ing members  of  (1) ;  then 

3  <2|^  4-  3  tu"  +  u^  =  19656  cu.  in.  (2) 

What  is  the  remainder  the  volume  of?  (See  the 
figures  of  the  last  article.)  Observe  that  the  seven  addi- 
tions to  the  cube  of  the  tens  have  one  dimension,  u,  in 
common,  and  that  equation  (2)  may  be  written 

(3  «2  +  3  <w  +  u^u  =  19656  cu.  in.  (3) 

Suppose  the  seven  solids  to  be  laid  side  by  side,  forming 
one  solid.  The  area  of  its  base  would  be  3  i^  -|-  3  ^it  +  w^, 
its  height  would  be  u^  and  its  volume  would  be  the  prod- 
uct of  its  base  and  its  height,  or  (3  ^^  +  3  tu  +  ii?)u. 

How  is  the  height  of  a  rectangular  solid  found  when  its 
volume  and  the  area  of  its  base  are  given  ?  To  find  the 
height  u  we  are  obliged  to  use  3 1"^  as  the  area  of  the  base, 
since  the  whole  area  is  not  yet  known.  If  we  find  that 
this  gives  too  large  a  value  for  w,  we  take  one  less. 

19656  -J-  3  «2  =  19656  --  (3  x  30^)  =7+. 

By  trial  we  find  this  too  large ;  hence  take  w  =  6. 
Then 

(3  ^'4-3  <w-f  w>=  (3  X  302-f  3  X  30  X  6 +6^  X  6=19656. 

Hence  -^46656  =  30  +  6  =  36.  Ans, 


CUBE  KOOT 


113 


Formula  for  Cube  Root 


3t' 


+  3tu-\-u^ 


3f  +  3tu-^u^ 


3t\  +  3tu^-i-w' 
3eu-\-3tu^-^u^ 


3  f  is  called  the  trial  divisor. 

3t^  -{-3tu-\-  u^  is  called  the  complete  divisor. 

Solution  of  -y/46656  by  the  formula : 

46656  I  30  +  6  =  36 
27000 
3  X  30^  =  2700  ' 
3  X  30  X  6  =    540 
6^=      36 


3276 


19656 


19656 


The  formula  for  the  cube  may  be  written 

(f  +  uf  =  f3  +  (3  f2  4-  3  /^  +  u^u.  (4) 

The  first  term  in  the  parenthesis  in  the  right  member  is 
the  trial  divisor;  the  whole  expression  within  the  paren- 
thesis is  the  complete  divisor. 


1.  ^15625. 

2.  ^166375. 


EXAMPLES  45 

3.  -^10648. 

4.  ^912673. 


5.  ^42875. 

6.  ^474552. 


83.  By  multiplying  the  square  of  h  +  t  -^u  (Art.  78) 
by  the  first  power,  the  pupil  may  prove  for  himself  that 

{h -{- 1 -{- uy  =  h^  +  f -\- u^ -\-  other  terms. 

I 


114  ALGEBRAIC   ARITHMETIC 

That  is,  the  cube  of  a  number  of  three  figures  contains, 
among  other  parts,  the  cube  of  each  of  the  figures.  The 
same  is  true  of  any  number. 

Thus  the  cube  of  382.5 

.53  =  .125 

23.     =  008. 

8»   .    =     512      . 
3»      .     =27 


contains 


plus  other  parts. 


It  will  be  seen  from  this  that  if  the  complete  cube  of 
any  number  be  separated  into  groups  of  three  figures 
each,  commencing  at  the  decimal  point,  the  number  of 
groups  will  be  equal  to  the  number  of  figures  in  the 
root ;  and  the  cube  of  each  figure  of  the  root  will  lie  wholly 
within  the  corresponding  group. 

The  last  group  to  the  left  may  contain  only  one  or  two 
figures. 

To  find  the  cube  root  of  any  number,  separate  it  into 
groups  as  above  directed,  and  proceed  as  in  the  last 
article,  always  regarding  the  part  of  the  root  already 
found  as  so  many  tens  with  respect  to  the  next  figure  of 
the  root. 

If  the  last  group  to  the  right  of  the  decimal  point  is 
incomplete,  it  must  be  completed  by  annexing  ciphers. 
No  such  number  has  an  exact  cube  root.     Why  not  ? 

When  a  cipher  occurs  in  the  root,  annex  two  ciphers 
to  the  trial  divisor,  and  another  group  to  the  dividend. 

If  there  is  a  remainder  after  the  root  of  the  last  period 
is  found,  the  result  may  be  found  to  as  many  places  as 
desired  by  annexing  groups  of  ciphers. 


^      0^7  Si 

UNIVEr^SiTY 

OF 

iLiFORNVh> 


CUBE   ROOT 


115 


Ex.  1.    ^12812'904. 


12'812'904  [234 
8 


3  X  202  =  1200 

3  X  20  X  3  =    180 

32=       9 


1389 
3  X  2302  =  158700 
3  X  230  X  4  =  2760 
42=    16 


4812 


4167 


161476 


645904 


645904 


Ex.  2.  ^8710.37. 


8'710'.370  1  20.57  + 
8 

3  X  202  =  1200 

3  X  2002  =  120000 

3  X  200  X  5  =   3000 

52=    25 

710 
710370 

123025 

615125 

3  X  20502  =  12607500 

3  X  2050  X  7  =   43050 

72=     49 

95245000 
88554195 

11. 

</{in- 

12. 

^Hm- 

13. 

^2i. 

14. 

^h 

116  ALGEBRAIC   ARITHMETIC 

EXAMPLES   46 

3.  ^1030301.  7.    ^.091125. 

4.  ^4492125.  8.    -^.000097336. 

5.  ^1045678.375.         9.    -^39.4995. 

6.  ^4080.659192.       10.    ^1250.6894. 

15.  What  are  the  dimensions  of  a  cube  that  has  the 
same  volume  as  a  box  2  ft.  8  in.  long,  2  ft.  3  in.  wide, 
and  1  ft.  4  in.  deep  ? 

16.  How  many  square  feet  in  the  surface  of  a  cube 
whose  volume  is  91125  cu.  ft.  ? 

17.  What  is  the  length  of  the  inner  edge  of  a  cubical 
bin  that  contains  150  bu.  ?  (1  bu.  contains  2150.42 
cu.  in.) 

18.  What  is  the  depth  of  a  cubical  cistern  that  holds 
200  bbl.  of  water?  (34  gal.  =  1  bbl. ;  1  gal.  =  231 
cu.  in.) 

19.  Find  the  length  of  a  cubical  vessel  that  will  hold 
4000  gal.  of  water. 

20.  What  are  the  dimensions  of  a  cubical  box  contain- 
ing ^  as  much  as  one  whose  edge  is  4  ft.  ? 


CHAPTER   VII 
MENSURATION 

84.  The  process  of  measuring  lines,  surfaces,  and  solids 
is  called  mensuration. 

Note.  All  the  rules  and  formulas  of  mensuration  and  the  facts 
upon  which  they  depend  are  proved  in  geometry.  When  state- 
ments are  made  without  explanation  in  this  chapter,  it  is  not  because 
none  can  be  given,  but  because  they  cannot  be  understood  without 
a  knowledge  of  geometry. 

85.  Lines.  A  straight  line  has  the  same  direction  at 
every  point. 

A  curved  line,  or  curve,  changes  its  direction  at  every 
point. 

Parallel  lines  have  the  same  direction ;  they  are  every- 
where equidistant. 

Two  straight  lines  are  said  to  be  perpendicular  to  each 
other  when  the  angle  between  them  is  a  right  angle 
(Art.  ^Q). 


straight  lines      — *■  ■• —  Parallel  lines 


Perpendiculftrs 
Curves 


117 


118  ALGEBRAIC   ARITHMETIC 

86.  Angles.  An  angle  is  the  difference  in  the  direction 
of  two  straight  lines. 

The  lines  are  called  the  sides  of  the  angle;  and  their 
point  of  meeting,  its  vertex. 

Angles  are  measured  in  degrees,  a  degree  being  ^^  of 
the  whole  angular  magnitude  about  a  point.  Thus  the 
sum  of  all  the  angles  that  can  be  drawn  with  a  common 
vertex  at  a  point  is  360°. 

If  two  lines  intersect  so  as  to  form  four  equal  angles, 
each  of  these  angles  is  a  right  angle.  A  right  angle  is  \ 
of  the  angular  magnitude  about  a  point,  and  is  therefore 
equal  to  90°. 

An  obtuse  angle  is  greater  than  a  right  angle. 

An  acute  angle  is  less  than  a  right  angle. 


1 


/ 


Acute  Right  Obtuse 

PLANE  FIGURES 

87.  A  portion  of  a  plane  surface  bounded  by  straight 
lines  or  curves  is  called  a  plane  figure.  The  sum  of  the 
lines  bounding  the  figure  is  called  its  perimeter. 

88.  Polygons.  Any  plane  figure  bounded  by  straight 
lines  is  called  a  polygon.  The  parts  of  a  polygon  are  its 
sides,  angles,  and  vertices. 

A  diagonal  of  a  polygon  is  a  straight  line  joining  any 
two  vertices  not  adjacent. 
Polygons  receive  special  names  according  to  the  num- 


PLANE  FIGURES 


119 


ber  of  their  sides.     A  triangle  has  three  sides ;  a  quadri- 
lateral, four ;  a  pentagon,  five ;  a  hexagon,  six. 

A  regular  polygon  has  equal  sides  and  equal  angles. 


Regular  Polygons 


Equilateral 
triangle 


Square 


Pentagon 


Hexagon 


o 


Octagon 


89.  Quadrilaterals  are  classified  as  follows : 


.  parallelo- 
gram has 
its  oppo- 
site sides 
parallel  (4 
classes). 


^  \ 


A   rectangle   has   four  right 
angles. 


A  square  is  a  rectangle  whose 
sides  are  equal. 


A  rhomboid  has  no  right  angles. 


A  rhombus  has  four  equal  sides 
and  no  right  angles. 


A  trapezoid  has  only  two  parallel  sides. 


A  trapezium  has  no  parallel  sides. 


7 


^ 


120 


ALGEBRAIC   ARITHMETIC 


90.   Area  of  Parallelograms. 

The  dimensions  of  a  parallelogram  are  its  base  (b)  and 
its  altitude  (a). 

Any  side  of  a  parallelogram  may  be  taken  as  its  base. 
Its  altitude  is  the  perpendicular  distance  between  its  base 
and  the  opposite  side. 


The  altitude  of  a  rectangle  is  equal  to  the  side  not 
taken  as  base. 

The  area  (A)  of  a  parallelogram  is  equal  to  the  product 
of  its  base  and  its  altitude. 

Explanation.  This  is  a  familiar  fact  in  the  case  of  rectangles  ; 
the  common  form  of  statement  for  this  case  being  that  the  area  of 
a  rectangle  is  equal  to  the  product  of  its  length  and  width. 

From  any  rhomboid  or  rhombus,  a  rectangle  of  the  same  dimen- 
sions can  be  constructed,  by  cutting  off  the  triangular  portion  from 
one  end  and  fitting  it  on  to  the  other,  as  shown  in  the  figure.  This 
change  in  the  form  of  the  figure  does  not  change  its  area,  since  the 
figure  is  composed  of  the  same  parts  as  before,  only  differently 
placed.  It  is  clear  that  the  area  of  the  figure  is  now  the  product 
of  its  two  dimensions  ;  hence  it  was  equal  to  the  product  of  these 
dimensions  before  its  form  was  changed. 

Definition.  Figures  having  the  same  area  are  called 
equivalent  figures. 

It  follows  from  what  has  been  said  above  that  paral- 
lelograms having  equal  bases  and  equal  altitudes  are 
equivalent. 


TRIANGLES 


121 


91.   Area  of  Triangles. 

The  dimensions  of  a  triangle  are  its  base  and  its  alti- 
tude. 

Any  side  of  a  triangle  may  be  taken  as  the  base.  The 
altitude  is  the  length  of  the  perpendicular  from  the 
base  to  the  opposite  vertex. 

The  area  of  a  triangle  is  equal  to  one-half  the  product 
of  its  base  and  altitude. 

Explanation.  Add  to  the  given  triangle  an  equal  triangle 
inverted.  The  two  together  form 
a  parallelogram  having  the  same 
base  and  altitude  as  the  given  tri- 
angle. Since  the  area  of  the  par- 
allelogram is  &a,  the  area  of  the 
triangle  is  ^  ha. 

It  follows  that  triangles  having  equal  bases  and  equal 
altitudes  are  equivalent. 


4  =  ^  6a 


92.  A  triangle  having  one  right  angle  is  called  a  right 
triangle. 

The  side  opposite  the  right  angle  is  called  the  hypothe- 
nuse,  and  the  other  two  sides,  the  legs. 

If  a  right  triangle  be  constructed  having  legs  3  and 

4  units  in  length  respectively,  the 
hypothenuse  will  be  found  to  be 

5  units  long ;  hence  the  sum  of  the 
areas  of  the  squares  constructed 
on  the  legs  will  be  equal  to  the 
area  of  the  square  constructed  on 
the  hypothenuse.  It  can  be  proved 
that  this  relation  is  true  of  any 
right  triangle.     That  is : 


122  ALGEBRAIC   ARITHMETIC 

The  square  of  the  hypothenuse   of  a  right  triangle  is 
equal  to  the  sum  of  the  squares  of  the  other  two  sides. 

This  relation  is  expressed  by  the  for- 
mula h^  =  6-  -f  a^,  where  a  and  b  are  the 
legs  of  the  right  triangle  and  h  is  the 
hypothenuse.      Subtract   a^  from   both 
sides  and  interchange  members;    then 
f,  —  y/h-i  —  a^         ^^  =  ^^^  ~  ^'^-      Take  the  square  root  of 
a  =  Vh'^  —  6^         both  sides,  and  we  have  b  =  V/i^  —  a\ 
In  the  same  way  we  obtain  a=  -Vh^  —  b\ 
If  any  two  sides  of  a  right  triangle  are  given,  the  third 
side  can  be  found  from  these  formulas. 


EXAMPLES  47 

1.  The  base  of  a  rhombus  is  10  ft.  6  in.,  and  its  alti- 
tude 8  ft.    What  is  its  area  ? 

2.  How  many  acres  in  a  piece  of  land  in  the  form  of 
a  rhomboid,  the  base  being  8.75  chains,  and  the  altitude 
6  chains  ? 

3.  Find  the  area  of  a  triangle  whose  base  is  12  ft. 
6  in.,  and  altitude  6  ft.  9  in. 

4.  What  is  the  cost  of  a  triangular  piece  of  land  whose 
base  is  15.48  chains,  and  altitude  9.67  chains,  at  $60 
an  acre? 

5.  Find  the  area  of  the  gable  end  of  a  house  that  is 
28  ft.  wide,  and  the  ridge  of  the  roof  15  ft.  higher  than 
the  foot  of  the  rafters. 

6.  What  is  the  base  of  a  triangle  whose  area  is  189 
sq.  ft.,  and  altitude  14  ft.  ? 


PROBLEMS  128 

7.  Find  the  altitude  of  a  triangle  whose  area  is  20J 
sq.  ft.,  and  base  3  yd. 

8.  The  legs  of  a  right  triangle  are  12  in.  and  16  in. 
respectively.     What  is  the  length  of  the  hypothenuse  ? 

9.  The  foot  of  a  ladder  is  15  ft.  from  the  base  of 
a  building,  and  the  top  reaches  a  window  36  ft.  above 
the  base.     What  is  the  length  of  the  ladder  ? 

10.  Find  the  distance  diagonally  across  a  floor  30 
by  40  ft. 

11.  What  is  the  length  of  a  path  diagonally  across 
a  10-acre  square  field  ? 

12.  A  room  is  20  ft.  long,  16  ft.  wide,  and  12  ft.  high. 
What  is  the  distance  from  one  of  the  lower  corners  to 
the  opposite  upper  corner? 

13.  The  hypothenuse  of  a  right  angle  is  35  ft.,  and 
one  leg  28  ft.     Find  the  other  leg. 

14.  A  ladder  52  feet  long  stands  against  the  side  of 
a  building.  How  many  feet  must  it  be  drawn  out  at  the 
bottom  to  lower  the  top  4  ft.  ? 

15.  Find  the  diagonal  of  a  cube  containing  729  cu.  in. 

16.  Wliat  is  the  side  of  a  square  field  whose  diagonal 
is  15  rods  ?    What  is  its  area  ? 

17.  A  ladder  28  ft,  long,  placed  in  a  street,  reaches 
the  top  of  a  building  18  ft.  high  on  one  side,  and  one 
15  ft.  high  on  the  other.     How  wide  is  the  street? 

18.  Two  vessels  sail  from  the  same  point,  one  north 
58  miles,  and  the  other  west  72  miles.  How  far  apart 
are  they  ? 


124 


ALGEBRAIC   ARITHMETIC 


93.   Area  of  a  Trapezoid. 

The  parallel  sides  of  a  trapezoid  are  called  the  bases ; 
and  the  distance  between  them  is  the  altitude. 

The  area  of  a  trapezoid  is  equal  to  one-half  the  product 
of  the  altitude  and  the  sum  of  the  bases. 

Explanation.  Draw  a  diagonal.  This  divides  the  trapezoid 
into  two  triangles  whose  common  altitude  is 
the  altitude  of  the  trapezoid.  The  base  of  one 
of  the  triangles  is  the  lower  base  of  the  trap- 
ezoid, the  base  of  the  other  is  the  upper  base 
of  the  trapezoid. 

The  area  of  the  one  triangle  is  }  abi,  of  the 
other  is  ^  abz  ;  the  area  of  the  trapezoid  is  the 


4  =  ^0(61  +  62) 
sum  of  the  areas  of  the  triangles.     Hence 


^  =  I  a6i  +  ^  afta  =  i  a(&i  +  &2). 

Note.  The  subscripts  (1  and  2)  used  here  and  in  subsequent 
articles  have  no  numerical  signification,  but  are  merely  used  to 
distinguish  between  two  values  of  the  same  letter. 

94.  The  area  of  a  trapezium  can  be  found  as  the  sum  of 
the  areas  of  two  triangles,  if  the  length  of  a  diagonal  and 


of  the  two  perpendiculars  from  it  to  the  opposite  vertices 
are  known. 

The  area  of  any  polygon  can  be  found  by  dividing  it 
into  triangles,  computing  their  areas  separately,  and  add- 
ing the  results. 


POLYGONS 


126 


95.  Area  of  Regular  Polygons.     Every  regular  polygon 
can    be    divided    into    equal    triangles 
having  a  common  vertex  at  the  centre 
of  the  polygon. 

The  altitude    of    these    triangles  is 
called  the  apothem  of  the  polygon. 

In  the  figure  the  dotted  line  is  the 
apothem. 

Since  the  area  of  the  polygon  is  the  sum  of  the  areas 
of  these  equal  triangles,  it  follows  that : 

The  area  of  a  regular  polygon  is  equal  to  one-half  the 
product  of  its  perimeter  (p)  and  apothem  (a). 


EXAMPLES  48 

1.  Eind  the  area  of  a  trapezoid  whose  bases  are  23  ffe. 
and  11  ft.,  and  the  altitude  9  ft. 

2.  One  side  of  a  quadrilateral  field  measures  38  rd., 
the  side  opposite  and  parallel  to  it  measures  26  rd.,  and 
the  distance  between  the  two  sides  is  10  rd.  Find  the 
area. 

3.  Find  the  area  of  a  trapezium  whose  diagonal  is 
42  ft.,  and  the  perpendiculars  to  this  diagonal  from  the 
opposite  vertices  are  16  ft.  and  18  ft. 

4.  Derive  the  formula  a  =  -  V3  for  the  altitude  (a)  of 

an  equilateral  triangle,  the  length  of  whose 
sides  is  b. 

5.  Find  by  the  formula  of  Ex.  4  the  alti- 
tude of  an  equilateral  triangle  whose  sides 
are  9  in. 


126  ALGEBRAIC   ARITHMETIC 

6.  Find  the  area  of  a  regular  hexagon  whose  sides  are 
10  in. 

Suggestion.  A  regular  hexagon  is  divided  into  six  equilateral 
triangles  by  its  three  diagonals  passing  through  the  centre.  Hence 
the  apothem  can  be  found  by  the  formula  of  Ex.  4. 

96.  The  Circle.  A  circle  is  a  plane  figure  bounded 
by  a  curve,  called  the  circumference,  all 
points  of  which  are  at  an  equal  dis- 
tance from  a  point  within  called  the 
centre. 

The  radius  (r)  of  a  circle  is  the  dis- 
tance from  its  centre  to  its  circumfer- 
ence (c).   Its  diameter  (d)  is  the  distance 
across  it  measured  through  the  centre.     Hence   d  =  2r. 

97.  The  ratio  of  the  circumference  of  a  circle  to  its  diam- 
eter is  the  same  for  all  circles,  and  is  generally  denoted  by 

the  Greek  letter  tt  (pronounced  pie).     That  is,  -  =  tt. 

The  value  of  tt  is  a  little  less  than  3f ;  more  accu- 
rately, 3.1416.  It  is  not  exactly  expressible  by  any  num- 
ber ;  but  can  be  found  to  as  many  decimal  places  as 
desired. 

From  the  equations  -  =7r  and  d  =  2  r  we  have 
d 

(?  =  w(/  =  2  irr,  (1) 

</=|,  (2) 

.  =  A.  (3) 


THE  CIRCLE  127 

Hence,  if  the  radius,  the  diameter,  or  the  circumfer- 
ence of  a  circle  is  given,  the  other  dimensions  can  be 
computed. 

98.  The  Area  of  a  Circle.  If  a  regular  polygon  of  any 
number  of  sides  be  circumscribed  about  a  circle,  its 
apothem  will  be  the  radius  of  the  circle. 


Let  p  denote  the  perimeter  of  the  polygon,  r  (radius 
of  circle)  its  apothem,  and  Ai  its  area ;  then 

A^==\rp.  (Art.  95) 

It  is  evident  that  Ai  is  larger  than  the  area  {A)  of 
circle,  and  that  p  is  larger  than  the  circumference  (c) 
of  the  circle. 

But  the  greater  the  number  of  sides  of  the  polygon, 
the  more  nearly  will  p  be  equal  to  c,  and  also  the  more 
nearly  will  Ai  be  equal  to  A. 

If  we  should  go  on  increasing  the  number  of  sides  of 
the  polygon,  its  area  would  still  be  found  by  the  formula 
Ai  =  ^  rp,  and  at  the  same  time  we  could  make  p  as 
nearly  equal  to  c  and  Ai  as  nearly  equal  to  A  as  we 
please. 

Hence  it  follows  that  it  must  at  least  be  very  nearly 


128  ALGEBRAIC    ARITHMETIC 

correct  to  find  the  area  of  the  circle  by  the  same  formula. 
It  is  proved  in  geometry  that  it  is  exactly 
correct. 

Hence  A  =  ^cr.  (1) 

This  amounts  to  regarding  the  circle  as 

composed  of  a  very  great  (infinite)  num- 

-  _        _     2     ber  of  triangles,  whose  common  altitude 

is  the  radius  of  the  circle,  and  the  sum 

of  whose  bases  is  the  circumference. 

Since  c  =  2  Trr,  (1)  may  be  written 

>f  =  u/^;  (2) 

which  is  the  usual  formula  for  finding  the  area  of  a  circle. 
It  is  sometimes  convenient  to  use  the  formula 

^=iw^  (3) 

which  the  pupil  may  derive  for  himself  from  (1)  and  the 
equation  d  —  2r. 

Exercise.     Give  the  meaning  of  formulas  (1),  (2),  and 
(3)  in  words. 

99.   Let  Ci  and  Ca  denote  the   circumferences  of  two 
circles ;  r^  and  r^  their  radii. 

Then  Cj  =  2  irri,  Cj  =  2  ^r^.  [Art.  97  (1)] 

Hence,  dividing  the  members  of  the  first  equation  by 
the  corresponding  members  of  the  second, 

£i  =  ?Z!i  =  !i;  (1) 

Ca     2  ttTj     rg ' 

or,  in  words :   The  ratio  of  the  ci/rcumferences  of  two  circles 


THE   CIRCLE  129 

is  equal  to  the  ratio  of  their  radii.     Or,  more  briefly :  The 
circumferences  of  two  circles  are  to  each  other  as  their  radii. 

Exercise.  Prove  that  the  circumferences  of  two  cir- 
cles are  to  each  other  as  their  diameters. 

100.  Let  Ai  and  A2  be  the  areas  of  two  circles ;  rj  and 
r2  their  radii. 

Then  A^  =  ttTj",  A^  =  Trr/.  [Art.  98  (2)] 

That  is :  The  ratio  of  the  areas  of  two  circles  is  equal  to 
the  ratio  of  the  squares  of  their  radii.  Or,  The  areas  of  two 
circles  are  to  each  other  as  the  squares  of  their  radii. 

Taking  the  square  root  of  the  first  and  last  members 
of  (1),  and  interchanging  them,  we  have 

In  words :  The  radii  of  two  circles  are  to  each  other  as 
the  square  roots  of  their  areas. 

Note.  The  relation  between  (1)  and  (2)  is  expressed  by  say- 
ing that  either  is  the  converse  of  the  other. 

Exercise.  (1)  Prove  from  Art.  98  (3)  that  the  areas 
of  two  circles  are  to  each  other  as  the  squares  of  their 
diameters.     Prove  the  converse. 

(2)  Prove  from  Art.  98  (1)  that  the  areas  of  two  circles 
are  to  each  other  as  the  squares  of  their  circumferences. 
Prove  the  converse. 


130  ALGEBRAIC   ARITHMETIC 

EXAMPLES  49 

1.  What  is  the  circumference  of  a  circle  whose  diame- 
ter is  20  in.  ? 

2.  What  is  the  diameter  of  a  tree  whose  girt  is  18  ft. 
6  in.? 

3.  Find  the  area  of  a  circle  whose  diameter  is  10  ft. 

4.  The   distance  around  a  circular  park  is   l}j  mi. 
How  many  acres  does  it  contain? 

5.  What  is  the  circumference  of  a  circle  whose  area 
is  19.635  sq.  ft.  ? 

6.  What  is  the  side  of  a  square  inscribed  in  a  circle 
whose  diameter  is  6  rd.  ? 

7.  The  area  of  a  circle  is  78.54  sq.  ft.     Find  the  side 
of  the  inscribed  square. 

8.  What  is  the  circumference  of  a  circular  pond 
whose  radius  is  11  rd. ?     Its  area? 

9.  What  is  the  radius  of  a  circle  equal  in  area  to  a 
triangle  whose  base  is  13  ft.  and  altitude  10  ft.  ? 

10.  A  cow  is  one  day  tied  to  the  top  of  -a  stake  5  ft. 
high  by  a  rope  20  ft.  long.  On  the  next  day  she  is  tied 
to  the  bottom  of  the  stake  by  the  same  rope.  Find  the 
difference  in  the  areas  over  which  she  can  graze. 

11.  What  will  it  cost  at  $  2  a  rod  to  fence  a  circular 
plot  of  land  containing  1  acre  ? 

12.  How  many  times  will  a  carriage  wheel  4  ft.  in 
diameter  turn  round  in  going  1  mi.  ? 

13.  A  square  field  contains  31.5  acres.  What  is  the 
length  of  its  diagonal  ?  What  is  the  circumference  of  a 
circular  field  of  the  same  area  ? 


SIMILAR   PLANE  FIGURES  131 

101.  Similar  plane  figures  are  plane  figures  having  the 
same  shape ;  that  is,  their  corresponding  angles  are  equal 
and  their  corresponding  lines  (like  dimensions)  are  pro- 
portional. 

Similar  figures  may  be  regarded  as  enlarged  or  reduced 
copies  of  one  another. 

All  circles  are  similar  figures,  and  all  regular  polygons 
of  the  same  number  of  sides. 

It  is  proved  in  geometry  that : 

(i.)  Any  corresponding  lines  of  similar  plane  figures  are 
to  each  other  as  their  other  corresponding  lines. 

(ii.)  The  areas  of  similar  plane  figures  are  to  each  other 
as  the  squares  of  their  corresponding  lines. 

Conversely, 

(iii.)  The  corresponding  lines  of  similar  plane  figures  are 
to  each  other  as  the  square  roots  of  their  areas. 

Note.  These  general  truths,  or  theorems,  were  proved  in 
Arts.  99  and  100  for  circles.  Compare  carefully  the  theorems  as 
given  for  circles  with  the  more  general  corresponding  theorems  of 
this  article. 

EXAMPLES  50 

1 .  The  length  of  one  side  of  a  triangular  field  contain- 
ing 2  A.  80  sq.  rd.  is  12  chains.  Find  the  area  of  a  field 
of  similar  shape  whose  corresponding  side  is  48  chains. 

Suggestion.     122  :  482  :  :  2.5  A. :  x  A.  [Theorem  (ii.)] 

2.  The  side  of  a  square  field  containing  18  acres  is 
60  rd.  long.  Find  the  side  of  a  square  field  that  contains 
J  as  many  acres. 


132  ALGEBRAIC   ARITHMETIC 

3.  Two  circles  are  to  each  other  as  9  to  16,  the  diame- 
ter of  the  less  being  112  ft.  What  is  the  diameter  of  the 
greater  ? 

SoGGESTiON.     3:4::  112  :  oc.  [Theorem  (iii.)] 

4.  A  rectangular  field  contains  720  sq.  rd.,  and  its 
length  is  to  its  breadth  as  5  to  4.  What  are  its  dimen- 
sions ? 

Suggestion.  Let  I  —  length  of  field  and  6  =  its  breadth.  The 
area  of  a  rectangle  5  by  4  is  20.     Hence 

20  :  720  :  :  52  :  Z2  ;    20  :  720  :  :  4^  :  h"^. 
Solve  the  proportions  for  P'  and  6^^  then  extract  the  square  roots. 

5.  It  is  required  to  lay  out  283  A.  107  sq.  rd.  of  land 
in  the  form  of  a  rectangle  so  that  the  length  shall  be  3 
times  the  width.     Find  the  dimensions. 

6.  A  pipe  1.5  in.  in  diameter  fills  a  cistern  in  5  hr. 
Find  the  diameter  of  a  pipe  that  will  fill  the  same  cistern 
in  55.1  min. 

7.  If  it  costs  $  167.70  to  enclose  a  circular  field  con- 
taining 17  A.  110  sq.  rd.,  what  will  it  cgst  to  enclose 
another  \  as  large  with  the  same  kind  of  fence? 

8.  If  63.39  rd.  of  fence  will  enclose  a  circular  field 
containing  2  A.,  what  length  will  enclose  a  circular  field 
of  8  A.? 

SOLIDS 

102.  Prisms  and  Cylinders.  The  word  solid  as  used  in 
mathematics  means  a  portion  of  space  bounded  by  sur- 
faces.   It  has  no  reference  to  what  the  space  may  contain. 


PRISMS   AND  CYLINDERS 


133 


A  solid  whose  ends  are  equal  and  parallel  polygons 
and  whose  sides  are  rectangles  is  called  a  right  prism. 

The  height  of  a  prism  is  the  perpendicular  distance 
between  its  ends,  or  bases. 

From  the  form  of  their  bases  prisms  are  called  tri- 
angular, quadriangular,  pentagonal,  etc. 

A  right  prism  whose  bases  are  rectangles  is  called  a 
quadrangular  prism,  rectangular  solid,  or  parallelopiped. 

A  cube  is  a  rectangular  solid  whose  faces  are  all  equal 
squares. 


Pentagonal 
Prism 


Cylinder 


Note.    The  space  passed  through  by  a  moving  surface  is  called 
the  solid  generated  by  the  surface. 


f€E^ 


The  solid  generated  by  a  rectangle  rotating 
about  one  of  its  sides  is  called  a  right  circular 
cylinder. 

Note.  The  word  prism  is  often  used  for  right 
prism,  and  cylinder  for  right  circular  cylinder.  They 
are  so  used  in  what  follows. 


103.  The  area  of  the  lateral  surface  {S)  of  a  prism  or 
a  cylinder  is  equal  to  the  product  of  its  height  (h)  and  the 
perimeter  (p)  of  its  base. 


134 


ALGEBKAIC   ARITHMETIC 


The  volume  (V)  of  a  rectangular  solid  is  equal  to  the 
product  of  its  three  dimensions. 

The  volume  of  a  prism  or  a  cylinder  is  equal  to  the 
product  of  its  height  (h)  and  the  area  (A)  of  its  base. 


A       / 

/-T-A 

V  =  lbh 


-p 


r|    . 

kj 

/ 

5  =  2irr/» 

S  =  hp 

K  =  irr2A 

¥  =  hA 

EXAMPLES   51 

1.  Find  the  area  of  the  lateral  surface  of  a  prism 
whose  altitude  is  7  in.,  and  its  base  a  pentagon,  each  side 
of  which  is  4  in. 

2.  What  is  the  entire  surface  of  a  cylinder  formed 
by  the  revolution  about  one  of  its  sides  of  a  rectangle 
6  ft.  6  in.  long  and  4  ft.  wide  ? 

3.  Find  the  solid  contents  of  a  cylinder  whose  alti- 
tude is  15  ft.,  and  its  radius  1  ft.  3  in. 

4.  Find  the  entire  surface  of  a  prism  whose  base  is 
an  equilateral  triangle,  the  perimeter  being  18  ft.,  and 
the  height  15  ft. 

5.  Find  the  contents  of  a  box  whose  length,  width, 
and  depth  are,  respectively,  4  ft.,  3  ft.,  and  2  ft. 

6.  Find  its  surface. 


PYRAMIDS   AND  CONES  135 

7.  Find  the  number  of  square  feet  necessary  to  make 
a  stove  pipe  2^  ft.  long  and  5  in.  in  diameter. 

8.  Find  the  amount  of  tin  necessary  to  make  a  tin 
pail  cylindrical  in  form,  6  in.  in  diameter  and  8  in.  deep, 
without  a  cover. 

9.  How  many  quarts  will  the  pail  hold  ? 

10.  Find  the  depth  of  a  cylindrical  tank  that  holds 
20  gal.  and  is  18  in.  in  diameter. 

11.  A  rectangular  can  is  10  in.  square  on  the  bottom 
and  holds  5  gal.     How  deep  is  it  ? 

12.  What  is  the  difference  in  the  number  of  square 
feet  of  lumber  necessary  to  make  the  sides  of  a  room 
16  ft.  long,  12  ft.  wide,  and  10  ft.  high,  and  one  of  cir- 
cular floor  containing  the  same  area  and  of  the  same 
height? 

104.  Pyramids   and  Cones.     A  regular  pyramid  is  a 

solid  whose  base  is  a  regular  polygon,  and  whose  sides 
are  equal  triangles  which  terminate  in  a  common  vertex. 

The  common  altitude  of  the  triangular  sides  is  called 
the  slant  height  of  the  pyramid. 

The  solid  generated  by  a  right  triangle 
rotating  about  one  of  its  legs  is  called  a 
right  circular  cone. 

The  length  of  the  hypothenuse  of  the 
generating  triangle  is  the  slant  height  of 
the  cone. 

The  height  of  a  pyramid  or  cone  is  the  perpendicular 
distance  from  its  vertex  to  its  base. 


136 


ALGEBRAIC    ARITHMETIC 


The  words  pyramid  and  cone  are  frequently  used  for 
regular  pyramid  and  right  circular  cone  respectively. 


Pjrramid 


Frustum 


Cone 


The  frustum  of  a  pyramid  or  of  a  cone  is  the  part  that 
remains  after  cutting  oif  a  portion  of  the  top  by  a  plane 
parallel  to  the  base. 

105.  The  lateral  surface  {S)  of  a  pyramid  or  a  cone  is 

equal  to  one-half  the  product  of  its  slant  height  (a)  and  the 
perimeter  (p)  of  its  base. 

This  follows  directly  from  the  formula  of  Art.  95,  and 
formula  (1)  of  Art.  98. 


S  =  ^ap  =  irar 


The  lateral  surface  of  a  frustum  of  a  pyramid  or  a  cone 

is  equal  to  one-half  the  product  of  the  slant  height  (a)  and 
the  sum  of  the  perimeters  (pi  and  j9a)  of  its  bases. 


PYRAMIDS   AND   CONES 


13T 


This  follows  for  the  frustum  of  a  pyramid  from  Art. 
93.  The  lateral  surface  of  the  frustum  of  a  cone  may 
be  regarded  as  made  up  of  a  very  great  (infinite)  number 
of  trapezoids. 


S  =  tra(ri  +  r^) 


Let  Vi  be  the  radius  of  the  lower  base  of  the  frustum 
of  a  cone,  and  7*2  the  radius  of  the  upper  base. 

Then      S  =  ia(pi-\- p^)  =  ^ a(2  -rrr^  -\-  2 TrVi) 
=  1  a  X  2 TT  (ri  4-  ^2)  =  7ra(ri  +  r^). 

The  volume  of  a  pyramid  or  a  cone  is  equal  to  one-third 

the  product  of  its  height  (Ii)  and  the  area  (A)  of  its  base. 
(See  note  to  Art.  84.) 

The  volume  of  the  frustum  of  a  pyramid  or  a  cone  is 
found  as  follows :  Add  the  areas  (A^  and  A2)  of  the  bases 
and  the  square  root  of  their  product,  and  multiply  this  sum 
by  one-third  of  the  height.     (See  note  to  Art.  84.) 

For  the  frustum  of  a  cone, 


V=\h{A,-^A,+  ^A,A,) 


=  |/l  (Trri^  -f-  -nri  +  -^/irr^Trr^ 
=  \h (irri  +  irr^^  4-  Trrjra) 
=  i'irh(ri^-\-r^^  +  riri). 


138  ALGEBRAIC   ARITHMETIC 


EXAMPLES  52 


1.  Find  the  lateral  surface  of  a  triangular  pyramid, 
the  slant  height  being  16  ft.,  and  each  side  of  the  base 
5  ft. 

2.  Find  the  lateral  surface  of  a  cone  whose  diameter 
is  17  ft.  6  in.,  and  the  slant  height  30  ft. 

3.  Find  the  entire  surface  of  a  square  pyramid  whose 
base  is  8  ft.  6  in.  square,  and  its  slant  height  21  ft 

4.  How  many  cubic  feet  in  the  mast  of  a  ship,  its 
height  being  50  ft.,  the  circumference  at  one  end  5  ft., 
and  at  the  other  3  ft.  ? 

5.  Find  how  much  water  can  be  put  into  a  tin  pail 
10  in.  deep,  like  a  frustum  of  a  cone  in  form,  whose  bot- 
tom is  8  in.  across,  and  top  12  in.  across. 

6.  How  many  square  feet  of  tin  in  the  pail  described 
in  the  last  example,  without  cover  ? 

7.  A  conical  wood  pile  is  6  ft.  high  and  12  ft.  in 
diameter  at  the  base.     How  many  cords  are  in  it? 

8.  How  many  bushels  of  oats  in  a  conical  pile  2  ft. 
high  and  12  ft.  around  it  at  the  base  ? 

9.  Find  the  number  of  cubic  feet  enclosed  by  a  barn 
60  ft.  long,  40  ft.  wide,  and  20  ft.  high,  with  a  pyramidal 
roof  8  ft.  high ;  all  inside  measurements. 

10.  How  many  cubic  feet  of  wood  are  in  a  log  20  ft. 
long  and  14  in.  in  diameter  ? 

11.  At  28  cents  per  cubic  foot,  what  is  the  cost  of  a 
stone  wall  28  in.  thick  at  the  base  and  18  in.  at  the  top, 
4  ft.  high  and  36  id.  long? 


THE  SPHERE 


139 


12.  How  many  cubic  feet  in  a  regular  eight-sided  post 
10  ft.  high,  the  width  of  one  side  being  3  in.,  and  the  dis- 
tance through  it  7.24  in.  ? 

106.  The  Sphere.  A  sphere  is  a  solid  bounded  by  a 
uniformly  curved  surface,  all  points  of  which  are  equally 
distant  from  a  point  within  called  the  centre. 

A  sphere  is  generated  by  a  semicircle  rotating  about  its 
diameter.  The  radius  and  the  diameter  of  the  generating 
semicircle  are  the  radius  and  the  diameter,  respectively, 
of  the  sphere. 

The  section  of  a  sphere  made  by  a  plane  passing 
through  its  centre  is  called  a  great  circle  of  the  sphere. 

107.  The  surface  (S)  of  a  sphere  is  equal  to  the  lateral 
surface  of  the  circumscribed  cylinder.  (Art. 
84,  note.) 

The  diameter  and  the  height  of  the 
circumscribed  cylinder  are  each  equal  to 
the  diameter  of  the  sphere ;  hence 

^  =  2  Trr  X  2  r  =  4  Trr*. 

The  area  of  the  surface  of  a  sphere  is 
$  =  A  irr^,  equal  to  four  times  the  area  of  its  great 
'^  =  f  »/*         circle. 

108.  Let  ri  and  r,  be  the  radii  of  two  spheres.  Si 
and  S2  their  surfaces. 

Then  Si  =  4:7rr^%    S2  =  4:nri\ 


Hence 


§1 
S, 


4  7rn^ 


2» 


or 


7rr2      rj 
S,:S,::n':r,'. 


140  ALGEBEAIC   ARITHMETIC 

The  surfaces  of  two  spheres  are  to  each  other  as  the 
squares  of  their  radii. 

Conversely,  Li  =  :^i. 

The  radii  of  two  spheres  are  to  each  other  as  the  square 
roots  of  their  surfaces. 

Exercise.  (1)  Prove  that  the  surfaces  of  two  spheres 
are  to  each  other  as  the  squares  of  their  diameters. 
Prove  the  converse. 

(2)  Prove  that  the  surfaces  of  two  spheres  are  to  each 
other  as  the  squares  of  the  circumferences  of  their  great 
circles.     Prove  the  converse. 

109.  The  Volume  of  a  Sphere.  We  have  seen  that  a 
circle  may  be  regarded  as  made  up  of  a  very  great  num- 
ber of  triangles  having  a  common  vertex  at  its  centre. 
Similarly,  a  sphere  may  be  regarded  as  made  up  of  a 
very  great  (infinite)  number  of  pyramids,  having  their 
bases  in  the  surface  of  the  sphere  and  their  common  ver- 
tex at  its  centre. 

The  surface  of  the  sphere  is  the  sum  of  the  bases  of 
these  pyramids,  and  its  radius  is  their  height. 

Now  the  volume  of  a  pyramid  is  the  product  of  its  base 
and  one-third  its  height;  hence  the  volume  of  a  sphere 
is  the  product  of  its  surface  and  one-third  its  radius. 

Hence,  since  ^S  =  4  irr'^, 

V=^\Sr  =  ^^'^.  (1) 

110.  Let  r^  and  r^  be  the  radii  of  two  spheres ;  Vx  and 
Fj  their  volumes. 


THE   SPHERE  141 


Then 

F,  =  i»n',   F,  =  tW. 

Hence 

V,     tirr/     ri' 

and 

The  volumes 

of 

two  spheres  are  to  each  other  as  the  cubes 

of  their  radii. 

Conversely, 

TJie  radii  of  two  spheres  are  to  each  other  as  the  cube  roots 
of  their  volumes. 

Exercise.  (1)  Prove  that  the  volumes  of  two  spheres 
are  to  each  other  as  the  cubes  of  their  diameters ;  as  the 
cubes  of  the  circumferences  of  their  great  circles. 

(2)  Prove  the  converse  of  each  of  the  above. 

EXAMPLES   53 

1.  Find  the  surface  of  a  sphere  whose  diameter  is  9  in. 

2.  Find  the  volume  of  a  sphere  whose  diameter  is  18  in. 

3.  The  glass  tank  of  a  lamp  is  spherical  in  shape  and 
4  in.  in  diameter  on  the  inside.  -  How  much  oil  will  it 
hold? 

4.  The  diameter  of  the  earth  is  about  8000  mi.  Find 
its  surface  and  volume. 

111.  Similar  solids  are  solids  having  the  same  form. 
Their  corresponding  surfaces  are  similar,  their  corre- 
sponding angles  equal,  and  their  corresponding  lines  pro- 
portional. 

All  spheres  are  similar  solids,  and  all  cubes, 


142  ALGEBRAIC   ARITHMETIC 

Theorems : 

(i.)  Any  corresponding  lines  of  similar  solids  are  to  each 
other  as  their  other  corresponding  lines. 

(ii.)  The  surfaces  of  similar  figures  (plane  or  solid)  are 
to  each  other  as  the  squares  of  their  corresponding  lines. 
Conversely, 

(iii.)  Tlie  corresponding  lines  of  similar  figures  are  to 
each  other  as  the  square  roots  of  their  surfaces. 

(iv.)  The  volumes  of  similar  solids  are  to  each  other  as 
the  cubes  of  their  corresponding  lines.     Conversely, 

(v.)  The  corresponding  lines  of  similar  solids  are  to  each 
other  as  the  cube  roots  of  their  volumes. 

Note.  These  theorems  were  proved  in  Arts.  108  and  110  for 
spheres.  Compare  the  theorems  given  there  with  the  more  general 
ones  of  this  article.     Why  are  these  more  general  ? 

The  reason  for  the  truth  of  these  theorems  is  that  all  lines  have 
but  one  dimension  —  length;  all  surfaces  are  proportional  to  the 
product  of  two  dimensions  —  length  and  width  ;  and  all  volumes 
to  the  product  of  three  dimensions  —  length,  width,  and  thickness. 
(Compare  with  Art.  68.) 

This  is  evident  in  the  case  of  squares  and  cubes.  If  the  side  of 
one  square  is  twice  that  of  another,  its  area  is  4  times  as  great. 
If  the  edge  of  one  cube  is  twice  that  of  another  cube,  its  volume  is 
8  times  as  great,  etc. 

Illustrate  the  last  two  statements  by  drawings. 

EXAMPLES  54 

1.  If  a  marble  column  10  in.  in  diameter  contains  27 
cu.  ft.,  what  is  the  diameter  of  a  column  of  equal  length 
that  contains  i81  cu.  ft.  ? 

2.  A  ball  4.6  in.  in  diameter  weighs  18  oz.    What  is 


SIMILAR   SOLIDS  143 

the  weight  of  another  ball  of  the  same  density,  that  is 
9  in.  in  diameter  ? 

3.  Two  vessels  have  the  same  shape.  One  is  12  in. 
deep  and  holds  7  gal.  The  other  is  7  in.  deep;  what 
does  it  hold  ? 

4.  A  tank  was  made  of  20  sq.  ft.  of  sheet  iron,  and  a 
tank  of  the  same  shape  of  30  sq.  ft.  What  is  the  ratio 
of  their  capacities  ? 

Suggestion.     First  find  the  ratio  of  their  linear  dimensions. 

5.  What  is  the  ratio  of  the  corresponding  edges  of 
two  similar  rectangular  solids  whose  volumes  are  respec- 
tively 2.7  cu.  ft.  and  1.5  cu.  ft.  ? 

6.  What  is  the.  edge  of  a  cube  whose  entire  surface  is 
1050  sq.  ft.,  and  what  is  its  volume  ? 

7.  What  must  be  the  inner  edge  of  a  cubical  bin  to 
hold  1250  bu.  of  wheat  ? 

8.  How  many  gallons  will  a  cistern  hold  whose 
depth  is  7  ft.,  the  bottom  being  a  circle  7  ft.  in  diameter 
and  the  top  5  ft.  in  diameter  ? 

9.  What  is  the  value  of  a  stick  of  timber  24  ft.  long, 
the  larger  end  being  15  in.  square,  and  the  smaller  6  in., 
at  28  cents  a  cubic  foot  ? 

10.  The  surface  of  a  sphere  is  the  same  as  that  of  a 
cube,  the  edge  of  which  is  12  in.  Find  the  volume  of 
each. 


Examples  1 

2. 

4^,  top; 

10.   6f. 

1.   5. 

20;*,  ball. 

11.    18. 

2.   6. 

3. 

lOj^,  sister ; 

12.   90. 

3.   16. 

20^,  brother. 

13.    If 

4.  4. 

4. 

16  and  34. 

14.   33,V 

5.   0. 

5. 

21,  42,  63. 

6.   8. 

, 

6. 

32;^. 

7.   28. 

7. 

13ida. 

1.    8a2. 

8.   66. 

8. 

7  marbles,  H ; 

2.    16acd. 

9.   26. 

19  marbles,  F. 

3.    6a8&. 

1©.   16. 

9. 

13  and  49. 

4.   20  a^bK 

11.   66. 

10. 

13  yr.,  son ; 

5.    Sa&Vrc. 

13.   3. 

58  yr.,  father. 

6.    S5  cex^y. 

13.    18. 

11. 

48. 

7.   3a +  36. 

14.    3. 

12. 

18  and  24. 

8.    15a -66. 

15.   tVV 

13. 

6^  bu. ; 
14,4  bu. 

9.  a^-^a"^. 
10.  2a5-a*. 

Examples  2 

14. 

I  0 

11.  3a3  +  6a2-6a. 

1.   3. 

12.   a62c2  +  a26ca 

2.   1. 

Examples  4 

+  a^b^c. 

8.   3f 

13.    18. 

4.   2. 

1. 

38. 

14.    300. 

5.   tV 

2. 

8. 

15.  900. 

6.    66. 

3. 

10. 

16.    76. 

7.   6. 

4. 

73. 

17.   29. 

8.  ^^: 

5. 

163. 

18.   33. 

6. 

2^. 
238. 

19.   6. 

Examples  3         1 

7. 

20.   72. 

1.   4yr., 

James ; 

8. 

9. 

21.  20. 

12  yr. 

,  John. 

9. 

36. 
146 

22.    13. 

23. 

25. 

12. 

3  a&  +  62  _  3  ^2. 

2. 

27r/o; 

24. 

729. 

13. 

31^,  1st; 

3H%; 

25. 

5a  +  6. 

62^,  2d ; 

^%^ 

26. 

2  a6  +  ac. 

97^,  3d. 

6f%; 

27. 

2  be. 

14. 

4hr. 

28f%. 

28. 

6ab-7ac+2bc. 

15. 

16. 

3. 

28000. 

29. 

26-|c. 

16. 

A,  24  apples ; 

4. 

$5. 

30. 

22  a6  -  8  ac. 

B,  12  apples  ; 

5. 

60%. 

31. 

x^-\-2xy  +  2/2. 

C,  16  apples. 

6. 

2500  sheep. 

82. 

4a2  +  4a6  4-62. 

17. 

40  men ; 

7. 

$217.61. 

83. 

a*  +  2  a^c  +  c2. 

80  boys ; 

8. 

$40000. 

34. 

3a2+5a6+2  62. 

880  women. 

9. 

15%. 

35. 

2a2+3a6-5  62. 

18. 

$3000. 

10. 

39  yd. 

36. 

ai-ab-2  b\ 

19. 

40  and  60. 

11. 

$3000. 

37. 

a3_2a6  +  62. 

20. 

12  and  24. 

12. 

52%; 

38. 

4a;2_i2a;y+9y2. 

21. 

6yr. 

$45,760. 

39. 

a«  -  2  a3c2  +  c*. 

22. 

10  yr.,  son ; 

40. 

2a2-3a&  +  62. 

30  yr.,  father. 

Examples  13 

41. 

10a2-9a62 

23. 

A,  $30; 

+  2  6*. 

B,$15; 

4. 

$2760. 

42. 

a3+3a26+3a62 

C,  $20. 

5. 

$5.25. 

+  68. 

24. 

164  artillery ; 

6. 

$  12800. 

472  cavalry ; 

7. 

$  1300  cost. 

Examples  6 

564  Infantry. 

8. 

Neither. 

25. 

14  yr. ; 

9. 

$720. 

1. 

3. 

16  yr. ; 

10. 

$300. 

2. 

3  a. 

18  yr. 

11. 

$1437.60. 

8. 

4  a. 

26. 

7  and  8. 

12. 

23%. 

4. 

3a;y. 

27. 

18  yr.; 

13. 

23i%. 

6. 

10  a6. 

36  yr. 

14. 

$1.47. 

6. 

aHd^. 

15. 

$.17^. 

7. 

4a2  +  6a6. 

Examples  11 

16. 

$96. 

8. 

3  a6  -  2  c. 

17. 

24%. 

9. 

fx-J2/2. 

1. 

A,  $1400; 

18. 

11%. 

10. 

a 

B,  $480; 

19. 

2|%loss. 

C,  $453.60 

20. 

$1.94^; 

11. 

a2  +  2a6  +  62. 

D,  $260. 

16f%. 

ANSWERS 


147 


21.  $50; 
$58.82; 
$65. 

22.  4%  loss. 

Examples  14 

4.   $378.13. 
6.    $96.90. 

6.  $6400.76  invest- 

ment. 

7.  6%. 

8.  $600. 

9.  $1271.88. 

10.  $  15  com.  ; 
$  750  inv. 

11.  $2905; 
9|^^perlb. 

12.  $2920. 

13.  14f%. 

14.  $432. 

15.  $506.25; 
23750  lb. 

16.  5|%. 

17.  $2100. 

18.  $  82.11  com.  ; 
$9301.89      pro- 
ceeds. 

19.  $10623.44. 

20.  $45111.44; 
$225.56. 

Examples  15 
1.    $555.75. 


3. 


$256.50. 

$1080; 


7. 
8. 

9. 
10. 


$696.11. 
$666.90  ; 

$3.75. 

$786. 
$363.80; 

$20.81. 
$2.40. 

Examples  16 


1.  $217.50. 

2.  $37.50. 

3.  $738. 

4.  $58000. 

5.  $13600. 

7.  $424. 

8.  $  44  prem. ; 
$5456  loss. 

9.  2o/„. 

10.  $1840.50    pre- 

mium ; 
$79959.50  value. 

11.  $32000.^ 

12.  $3717.83. 

13.  $3168. 

14.  $5600. 

Examples  17 

1.  $88000. 

2.  $19072.16. 

3.  $.45. 

4.  U%,  rate; 
$95.25,  A's  tax. 

5.  .0228  tax  rate  ; 
$214.66. 


7. 

8. 

9. 
10. 
11. 


$407.20. 

$224.37. 

$103.13. 

$21.43. 

2  J  mills. 

$76.39. 


Examples  19 

1. 

$7212.50. 

2. 

$37.15. 

3. 

$11925. 

4. 

$151.88. 

5. 

$30.72. 

Examples  20 

1. 

$3902.40. 

2. 

$  78133.33. 

3. 

$208.33. 

4. 

$6553.60. 

6. 

$25372. 

6. 

48  bu. 

7. 

$1700,  Istyr.; 

$1785,  2d  yr. 

8. 

40f%. 

9. 

$40842. 

10. 

61788.6  lb. 

11. 

25f-%  nearly. 

12. 

12%. 

13. 

$2116.94. 

14. 

16^. 

16. 

$863.99. 

$68.21. 

Examples  22 

2.  $83.22; 
$  843.22. 

3.  $11.91; 
$191.41. 


148 


ANSWERS 


4. 

$57.20; 

Examples  24 

Examples  28 

$382.20. 

2.    $6.06. 

3.    Mat.  Oct.  30 ;  81 

6. 

$146.19; 

3.    $82.36. 

da.     term     of 

$904.94. 

4.   $10.96. 

dis.;    $940.14 

6. 

$  142.03  ; 

5.   $39.55. 

proceeds. 

$1166.28. 

6.   $106.99. 

4.   Apr.  8  ;  46  da.  ; 

7. 

$42.28; 

7.    $172.17. 

$917.21. 

$626.78. 

8.  $51.37. 

6.   Aug.  13 ;  64  da. ; 

8. 

$55.77; 

9.   $205.48. 

$690.04. 

$781.61. 

6.    $641.55. 

9. 

$83.02; 

7.   $1821.60. 

$470.97. 

Examples  26 

8.    $2548.53. 

10. 

$.17; 

9.    $56.69. 

$42.37. 

1.   $516.71. 

2.    $1000. 

Examples  29 

Examples  23 

3.   3  yr.  4  mo.  24  da. 

2.    $2688. 

1. 

$58.93. 

4.   6%. 

3.    $464.10. 

2. 

$  8.40. 

6.    $650.80. 

4.   $3808. 

3. 

$67.67. 

6.    7%. 

6.    $586.04. 

4. 

$159.75. 

7.   7  mo.  10  da. 

6.    $247.52. 

5. 

$67.09. 

8.   7%. 

Examples  30 

6. 

7. 

a 

$38.11. 
$30.81. 
$8.93. 
$3647.61. 

9.   $669.12. 
10.    $1403.08. 

2.    $900  simple; 
$  1035  annual ; 

8. 
9. 

Examples  27 

$1046.30  comp. 
3.   $428.76. 

10. 
11. 

$1066.36. 
$2010.42. 

2.    $917.43; 

4.    $189.15. 
6.   $311.64. 

12. 
13. 
14. 
15. 

$142.45. 
$1886.17. 
$263.83. 
$410.70. 

$82.57. 

3.  $43.65  in  favor 

of  dis. 

4.  $1137.61. 

6.  $100.32. 

7.  $41.99. 

8.  $245.77. 

9.  $53.38. 

10.  $1540.79. 

11.  $2357.79. 

16. 

17. 

$25.78. 
$165.50. 

5.  $838.26. 

6.  The  first  by  5^ 

18. 

$410.73. 

per  ton. 

19. 

$1936.60. 

7.    $2010.13. 

Examples  31 

20. 

$1120.69. 

8.    $147.06. 

2.    $576. 

21. 

$7.33. 

9.   $.92. 

3.    $98.33. 

ANSWERS 

149 

4.   $575.34. 

6. 

280  bu. 

9. 

A,  $131; 

5.    $284.79. 

6. 

$6428.57. 

B,  $393; 

6.    $601.08. 

7. 

220|  cd. 

C,  $262. 

8. 

$52.79. 

10. 

A,  $1800; 

Examples  32 

9. 

9  men. 

B,  $600; 

2.    $577.38. 

10. 

2|da. 

C,  $1200. 

3.    $99.88. 

11. 

10  ft.  2|  in. 

11. 

1st,  $357f; 

4.    $676.67. 

12. 

10  ft. 

2d,  $642f. 

5.    $285.99. 

13. 

546  bbl. 

12. 

$171.60; 

6.   $603.49. 

14. 

2080  lb. 

$257.40. 

7.   $1386.78. 

15. 

$100. 

8.    $1284.11. 

16. 

15  da. 

Examples  37 

9.    $162.25. 

17. 

432  mi. 

2. 

June  27, 1897. 

10.  $4408.21. 

11.  $523.43. 

1. 

Examples  36 

A's,  $100; 

3. 

4. 

May  5,  1896. 
June  23. 

Examples  34 

B's,  $150. 

5. 

6  mo. 

3.  $4.05. 

4.  44|bbl. 

5.  16  men. 

2. 

A's,  $3200; 
B's,  $1800; 

C's,  $  1400. 

6. 

7. 
8. 

5  yr.  20  da.  from 
date  of  last 
payment. 

Nov.  26. 

Mar.  7 ; 

$1178.01. 

6.  96  sheep. 

7.  $5355. 

8.  7  hr.  13^  min. 

3. 
4. 

A's,  $  1800  ; 
B's,  $3000. 
A,  $1710; 

9.   112^  mi. 

B,  $  870.20. 

10.   59|da. 

5. 

A,  $6000; 

Examples  43 

11.    $7320. 

B,  $8402.25; 

3. 

115. 

12.    $10958.90. 

C,  $5055.75; 

4. 

109. 

13.   90  horses. 

D,  $3042. 

5. 

997. 

14.    2yr.  6mo. 

6. 

A's,  $200; 

6. 

143.2. 

15.    3hr. 

B's,  $240; 

7. 

54.64. 

16.   20  men. 

C's,  $160. 

8. 

.036. 

17.    75  da. 

7. 

A,  $2400; 

9. 

14.0048+. 

18.    9  men. 

B,  $2666.67; 

10. 

1.5006+ . 

C,  $2933.33. 

11. 

7.625. 

Examples  35 

8. 

A,  $388.71  ; 

12. 

4.213+. 

3.   $240. 

B,  $249.17; 

13. 

103.9. 

4.   $498.08. 

C,  $112.12. 

14. 

1.5411 +. 

150 


ANSWERS 


16. 
16. 


If 


17. 

tV^. 

18. 

.91287  +  . 

19. 

5.6. 

20. 

8^ 

21. 

1008  ft. 

22. 

240.33  rd. 

23. 

52  rd. 

24. 

$187.20. 

25. 

80  X  40  rd 

26. 

101.2  rd. 

27. 

107.33  rd. 

28. 

182  sq.  rd. 

29. 

49  rows. 

3. 

4. 

5. 

6. 

7. 

8. 

9. 
10. 
11. 
12. 
13. 
14. 
15. 
16. 
17. 
18. 
19. 
20. 


Examples  46 

,   101. 
165. 
101.5. 
15.98. 
.45. 
.046. 
3.4056. 
10.77. 

M. 

1.42+. 
.7936. 
2  ft. 

12150  sq.  ft. 
5  ft.  8+  in. 
9  ft.  5.3+  in. 
8  ft.  1.4  in. 
3.17  ft. 


Examples  47 

1.  84  sq.  ft. 

2.  5|A. 

3.  42j3^  sq.  ft. 

4.  $449.07. 

5.  210  sq.  ft. 

6.  27  ft. 

7.  4^  ft. 

8.  20  in. 

9.  39  ft. 

10.  50  ft. 

11.  56.57  rd. 

12.  28  ft.  3.36  in. 

13.  21ft. 

14.  20  ft. 

15.  15.59.  in. 

16.  10.6  rd.; 
112.5  sq.  rd. 

17.  45.08  ft. 

18.  92.45  mi. 

Examples  48 


1.  153  sq.  ft. 

2.  2  A. 
714  sq.  ft. 
Given. 
7.794+  in. 
259.8+  sq.  in. 

Examples  49 

1.  5  ft.  2.83  in. 

2.  5  ft.  10.67  in. 
8.    78.54  sq.  ft. 

4.  114.59  A. 

5.  15.708  ft. 


8.  4.24  rd. 

7.  7.07+  ft. 

8.  69.12  rd.  ; 
380.13  sq.  rd. 

9.  4.55  ft. 

10.  78.54  sq.  ft. 

11.  $89.68. 

12.  420+  times. 

13.  100.399    sq.    rd. 

diagonal ; 
251.6  rd.circum. 

Examples  50 

1.  40  A. 

2.  34.64+  rd. 

3.  149  ft.  4  in. 

4.  30  rd.;  20  rd. 

5.  369  rd. ;  123  rd. 

6.  3.5  in. 

7.  $75. 

8.  126.78  rd. 

Examples  51 

1.  140  sq.  in. 

2.  263.89  sq.ft. 

3.  73.63  cu.  ft. 

4.  301.177  sq.  ft. 

5.  24  cu.  ft. 

6.  62  sq.  ft. 

7.  3.27^  sq.  ft. 

8.  179.07  sq.  in. 

9.  3.91  qt. 

10.  18.16  in. 

11.  11.66  in. 

12.  68.80  sq.  ft. 


Examples  52 

Examples  53 

4. 

.54433. 

1. 

2. 

3. 

120  sq.  ft. 
824.67  sq.  ft. 
429^  sq.  ft. 

1.  254.47  sq.  in. 

2.  3053.6+  cu.  in. 
8.    .58  qt. 

5. 
6. 

1.2164+. 
13.228  ft.  edge ; 
2315.03  cu.  ft. 

4. 
6. 
6. 

64.99  cu.  ft. 
795.871  cu.  in. 
257  sq.  ft. 

4.   201062400     sq. 
mi.; 
268083200000 

7. 
8. 

vol. 
11  ft.  7  in. 
1494.257  gal. 

7. 
8. 
9. 

1  cd.  98^  cu.  ft. 
6.14  bu. 
54400  cu.  ft. 

cu.  mi. 
Examples  54 

9. 
10. 

$5.46. 

1  cu.  ft.  vol.  of 
cube; 

10. 

21.38  cu.  ft. 

1.    14.42  in. 

1    cu.    ft.     659 

11. 

$1276.12. 

2.   9  1b. 

cu.  in.  vol.  of 

12. 

3+  cu.  ft. 

8.    1.389+  gal. 

sphere. 

^     OF  THE 

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By  J.  B.  LOCK, 

Author  of  "  Trigonometry  for  Beginners"  "  Elementary  Trigonometry,"  etc^ 

Edited  and  Arranged  for  American  Schools 

By   CHARLOTTE    ANGAS    SCOTT,   D.SC, 

Head  of  Math.  Dept.,  Bryn  Mawr  College,  Pa. 

16mo.    Cloth.    75  cents. 

"  Evidently  the  work  of  a  thoroughly  good  teacher.  The  elementary  truth,  that 
arithmetic  is  common  sense,  is  the  principle  which  pervades  the  whole  book,  and  no 
process,  however  simple,  is  deemed  unworthy  of  clear  explanation.  Where  it  seems 
advantageous,  a  rule  is  given  after  the  explanation.  .  .  .  Mr.  Lock's  admirable 
'  Trigonometry '  and  the  present  work  are,  to  our  mind,  models  of  what  mathematical 
school  books  should  be."  — '  The  Literary  World. 


FOR   MORE   ADVANCED    CLASSES. 

\  ARITHMETIC. 

By  CHARLES  SMITH,  M.A., 
Author  of  *'  Elementary  Algebra,"  "A  Treatise  on  Algebra^* 

AND 

CHARLES   L.  HARRINGTON,  M.A., 

Head  Master  of  Dr.  J.  Sach's  School  for  Boys,  New  York. 
1 6mo.    Cloth.    00  cents. 

A  thorough  and  comprehensive  High  School  Arithmetic,  containing  many  good 
examples  and  clear,  well-arranged  explanations. 

There  are  chapters  on  Stocks  and  Bonds,  and  on  Exchange,  which  arc  of  more 
than  ordinary  value,  and  there  is  also  a  useful  collection  of  miscellaneous  examples. 


THE   MACMILLAN   COMPANY, 

66  FIFTH  AVENUE,  NEW  YORK. 


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SEP  28  W25 


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